83 
R5 


QUALITATIVE 
ANALYSIS 

TEST  AND  MCLAUGHLIN 


TABLE  OF 


K 

NA 

XII4 

MQ 

CA 

SR 

BA 

Zx 

MN 

Co 

Ni 

AL 

Ci 

32.95 
3.9 

35.80 
5.42 

10° 
33.0 

55.81 
5.1 

73.19 
5.4 

51.09 
3.0 

37.24 

1.7 

203.0 
9.2 

!.-)().() 

50.0 

s. 

400 

N03 

30.34 
2.6 

83.97 
7.4 

200 

74.31 
4.0 

121.8 
5.2 

66.27 
2.7 

8.74 
0.33 

117.S 
4.7 

v.  s. 

45.47 
5.4 

50 

v.  s. 

S04 

11.11 
0.62 

16.83 
1.15 

76.0 

35.43 

2.8 

0.20 

0.015 

0.011 
0.036 

0.0,23 

0.041() 

58.12 

3.1 

43.3 

36.21 
4.0 

106.0 

36.1 

OH 

112.9 
18.0 

116.4 

21.0 

0.001 
0.032 

0.17 
0.02 

0.77 
0.063 

3.7 
0.22 

0.085 
0.045 

ins. 

ins. 

ins. 

ins. 

S 

v.  s. 

s. 

s. 

hyd. 

hyd. 

hyd. 

hyd. 

o.oji!) 

0.0,71 

•  MU52 
0.0472 

0.0838 
0.0442 

0.0336 
0.044 

hyd. 

e6, 

lox.o 
6.9 

19.39 
1.8 

100 

0.1 

0.01 

0.0013 
0.0813 

0.0011 
0.047 

0.0023 

0.0,11 

0.004 
0.033 

ins. 

ins. 



hyd. 

i>o4 

25.0 

20.0 

(sec) 
25.0 

0.015 
0.035 

ins. 

ins. 

ins. 

ins. 

68.1 

2.7 

<'i:<>., 

61.21 
3.30 

29.0 

73.0 
4.3 

0.4 
0.03 

0.12 
0.0(16 

o.o.{35 
0.0414 

C204 

30.27 
1.6 

3.34 
0.24 

4.2 

0.03 
0.0027 

0.0855 
0.0443 

0.0046 
0.0.?26 

o.oosr, 

0.0,3S 

•i 

0.0864 
0.044 



ins. 



ins. 

NOTE.    Upper  numbers  in  each  square  are  grams  of  anhydrous  salt  in  100  grams  of 
water  at  ordinary  temperature  (18°).    Lower  numbers  are  molar  solubilities. 


SOLUBILITIES 


CK 

ins. 

FE 

CD 

Bi 

Cu 

PB 

!!<;" 

As 

SB 

SN 

Ho1 

AG 

(ic) 
158.0 

140.0 

hyd. 

60.0 

1.49 
0.05 

7.4 

s. 

910 

270 

0.045 

0.068 

0.0313 
0.059 

s. 

143 

hyd. 

125 

51.66 
1.4 

hyd. 

s. 

213.4 

8.4 

s. 

(ous) 
60.0 

s. 

hyd. 

20.0 

0.0041 
0.0313 

hyd. 



18.8 

si.  s. 

0.55 
0.020 

ins. 

(ous) 
0.0007 
0.048 

0.00026 
0.0518 

ins. 

ins. 

0.01 
0.034 

0.0051 
0.0322 

ins. 

ins. 

0.01 
0.001 

hyd. 

0.0362 
0.047 

0.0313 
0.059 

0.0418 
0.0.35 

0.0434 
0.0635 

0.0492 
0.0536 

0.0513 
0.0754 

0.0452 
0.0521 

0.0347 
0.0552 

ins. 

0.0414 
0.0,55 

hyd. 

ins. 

ins. 

ins. 

0.031 
0.043 

ins. 

0.003 
0.0gl 

ins. 

0.0414 
0.0617 

0.0364 
0.0416 

0.042 
0.0.5 

ins. 

ins. 

0.0025 
0.0815 

| 

v.  s. 

0.0033 
0.0317 

0.0316 
0.0554 

0.0034 
0.0371 

Where  exact  figures  are  not  given,  s.  means  "soluble";  v.  s.,  "very  soluble" 
sl.s.,  "  slightly  soluble  ";  ins.,  "insoluble";  hyd.,  ff hydrolyzed." 


NOTES  ON 
QUALITATIVE  ANALYSIS 


BY 

LOUIS  AGASSIZ  TEST 

n 

AND 

H.  M.  MCLAUGHLIN 


GINN  AND  COMPANY 

BOSTON    •    NEW  YORK    •    CHICAGO    •    LONDON 
ATLANTA    •    DALLAS    •    COLUMBUS    •    SAN  FRANCISCO 


COPYRIGHT,  1919,  BY 
LOUIS  AGASSIZ  TEST  AND  H.  M.  MCLAUGHLIN 


ALL  RIGHTS  RESERVED 
319.6 


fltbenatum 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 

Every  instructor  will  be  governed  in  his  teaching  of  a 
subject  by  the  object  to  be  attained  from  the  course,  the 
time  which  can  be  devoted  to  the  work,  and  his  method  of 
presentation. 

Qualitative  analysis  has  a  place  in  a  beginning  course  in 
general  chemistry  only  as  a  means  of  reviewing  principles 
already  studied  and  fixing  them  in  the  mind  of  the  student 
by  applications. 

When  qualitative  analysis  must  be  given  in  the  first  year 
before  the  general  chemistry  is  completed,  it  is  frequently 
difficult  to  know  how  to  distribute  the  time  between  the  two 
subjects.  In  such  cases  the  instructor  must  be  guided  not 
only  by  the  time  he  can  devote  to  the  work  but  by  the  prepa- 
ration of  his  students.  Most  of  the  texts  on  qualitative  analysis 
are  based  on  the  assumption  that  the  student  has  completed  a 
course  in  general  chemistry,  and  are  not  adapted  to  students 
who  have  had  but  one  semester  of  general  chemistry. 

In  preparing  this  outline  for  their  own  students  the 
authors  have  had  in  mind  the  preparation  of  a  manual  which 
their  students,  having  had  only  one  semester  of  general 
chemistry,  could  cover  in  six  hours  a  week  for  the  fifteen 
weeks  available  for  the  course.  No  excuses  are  made  for  the 
book  other  than  that  the  authors  have  found  it  helpful  in 
their  work  at  Iowa  State  College  and  more  satisfactory  than 
any  they  have  been  able  to  secure. 

The  questions  given  at  the  end  are  intended  to  cover  the 
work  given  in  the  manual  and  to  teach  the  student  to  think 
for  himself.  Some  are  probably  too  difficult  for  the  average 

iii  • 

M123271 


iv  NOTES  ON  QUALITATIVE  ANALYSIS 

student  in  the  time  which  may  be  at  his  disposal,  but  such 
questions  may  easily  be  omitted  at  the  discretion  of  the 
instructor. 

In  preparing  this  manual  frequent  recourse  has  been  had 
to  various  larger  and  more  complete  texts,  and  the  authors 
feel  particularly  indebted  to  Professors  Stieglitz  and  Tread- 
well  and  Hall,  which  indebtedness  they  wish  to  acknowledge. 

Thanks  are  also  due  to  the  assisting  staff  of  Iowa  State 
College,  all  of  whom  have  given  much  help  in  trying  out 
these  notes  with  classes. 

Particularly  do  the  authors  wish  to  thank  Dr.  W.  A. 
Roberts,  who  wrote  much  of  the  original  notes,  Miss  Nellie 
Nay  lor,  and  Mr.  R.  E.  Kirk  for  many  helpful  suggestions. 


CONTENTS 

PAGE 

INTRODUCTION .     .     .  1 

SOLUTIONS,  DEFINITIONS,  PROPERTIES  AND  LAWS  OF      ....  2 

SOLUBILITY-PRODUCT  PRINCIPLE 17 

HYDROLYSIS 18 

OXIDATION  AND  REDUCTION 20 

SOLUBILITIES 24 

GENERAL  INSTRUCTIONS f  25 

CLASSIFICATION  OF  METAL  IONS 29 

THE  ALKALI  GROUP .30 

THE  ALKALINE  EARTH  GROUP 31 

NOTES  ON  ANALYSIS 37 

SCHEME  FOR  ANALYSIS  OF  THE  ALKALI  AND  ALKALINE  EARTH 

GROUPS 38 

THE  SULFIDE  GROUPS 40 

SEPARATION  OF  THE  COPPER  AND  ARSENIC  GROUPS       ....  47 
THE  IRON-ZINC  GROUP 

AMPHOTERIC  HYDROXIDES 49 

FORMATION  OF  COMPLEX  IONS 50 

SEPARATION  AND  IDENTIFICATION  OF  THE  MEMBERS  OF  .     .  55 

SCHEME  FOR  ANALYSIS 56 

THE  COPPER  GROUP 59 

SCHEME  FOR  ANALYSIS  OF  THE  COPPER  GROUP 61 

THE  ARSENIC  GROUP 62 

SCHEME  FOR  ANALYSIS  OF  THE  ARSENIC  GROUP 63- 

SCHEME  FOR   SEPARATION   OF   THE  ARSENIC  AND   THE  COPPER 

GROUPS 65 

THE  SILVER  GROUP 66 

SCHEME  FOR  ANALYSIS  OF  THE  SILVER  GROUP 68 

v 


vi  NOTES  ON  QUALITATIVE  ANALYSIS 

PAGE 

ANALYSIS  FOR  ACID  IONS 69 

PREPARATION  OF  THE  SOLUTION  FOR  ANALYSIS 70 

TESTS  FOR  ANIONS  (SO-CALLED  ACIDS) 71 

METHODS  OF  PROCEDURE 

FOR  THE  SEPARATION  AND  IDENTIFICATION  OF  METAL  IONS 
OF  THE  IRON-ZINC  GROUP  IN  THE  PRESENCE  OF  PHOSPHATES 

OR  OXALATES 72 

WITH  AN  UNKNOWN  SUBSTANCE 73 

FOR  ALLOYS ,     .     .  74 

FOR  BAKING  POWDERS 75 

QUESTIONS 76 

INDEX 01 

TABLE  OF  SOLUBILITIES Inside  front  cover 

PERIODIC  SYSTEM  OF  THE  ELEMENTS   .     .     .  Facing  back  cover 
INTERNATIONAL  ATOMIC  WEIGHTS  .    Inside  back  cover 


NOTES  ON 
QUALITATIVE  ANALYSIS 


INTRODUCTION 

Qualitative  analysis  treats  of  the  identification  of  sub- 
stances. In  a  general  qualitative  analysis  the  problem  may 
be  the  identification  of  a  compound,  an  element,  or  some 
radical  or  group  of  elements.  This  outline  deals  with  the 
identification  of  the  more  common  metals  and  a  few  of  the 
important  nonmetallic  radicals.  These  latter  occur  as  negative 
ions  in  acids.  A  substance  is  identified  by  some  property 
which  it  possesses.  The  more  characteristic  and  perceptible 
to  our  senses  this  property  is,  the  more  certain  is  the  identity 
of  the  substance,  and  if  more  than  one  such  property  can  be 
made  use  of,  so  much  more  positive  is  the  identification. 
In  general  we  make  use  of  physical  properties.  These  may 
sometimes^  be  observed  in  mixtures,  but  more  frequently  it 
is  necessary  to  separate  the  substance  from  most  or  all  other 
material  with  which  it  may  be  found.  Such  separations  are 
almost  always  brought  about  or  made  possible  by  chemical 
means.  While  physical  properties  mostly  serve  for  identifi- 
cation, it  is  frequently  necessary  to  make  use  of  chemical 
reactions  in  order  that  the  element  or  radical  may  enter  into 
some  combination  having  properties  sufficiently  distinctive 
for  identification.  Such  reactions  as  may  cause  a  change  in 
color,  formation  of  a  precipitate,  dissolving  of  a  precipitate, 

l 


42  NOTES  ON  QUALITATIVE  ANALYSIS 

formation  of  a  gas,  distinctive  odor,  etc.,  are  spoken  of  as 
tests.  A  few  examples  will  illustrate: 

If  to  a  solution  of  some  unknown  substance  a  few  drops 
of  a  solution  of  potassium  thiocyanate  is  added  and  a  deep 
blood-red  color  is  produced,  it  is  considered  proof  that  ferric 
iron  was  in  the  unknown  substance.  If  sodium  chloride  is 
held  in  a  colorless  flame  ^a  bright-yellow  color  is  imparted  to 
the  flame.  This  is  used  as  a  means  for  the  identification  of 
sodium.  If  strontium  chloride  is  held  in  a  colorless  flame  a 
bright-red  color  is  imparted  to  the  flame.  This  may  be  used 
as  a  means  for  the  identification  of  strontium.  The  forma- 
tion of  a  precipitate  under  certain  conditions  is  an  important 
means  of  identification.  If  to  a  solution  of  some  unknown 
substance  a  solution  of  sodium  chloride  is  added,  and  a  white 
precipitate  forms  which  is  insoluble  in  cold  water,  it  will 
probably  be  either  lead  chloride,  mercurous  chloride,  or  silver 
chloride,  lead,  mercury,  and  silver  being  the  only  ordinary 
substances  that  behave  in  this  manner.  If  this  precipitate 
dissolves  in  ammonium  hydroxide  the  substance  is  most 
probably  a  silver  salt.  If  it  turns  black  on  the  addition  of 
ammonium  hydroxide  the  substance  is  a  mercury  compound. 

Since  most  of  the  reactions  used  in  identification  are 
double  decomposition  reactions  taking  place  in  solution,  a 
brief  review  of  the  nature  of  solution  and  the  ionic  theory 
will  be  taken  up.  As  most  double  decomposition  reactions 
in  solution  are  reversible,  the  idea  of  equilibrium  will  play 
an  important  part  in  the  discussion. 

SOLUTIONS 

"A  solution  may  be  defined  as  a  body  of  homogeneous 
character  whose  composition  may  vary  between  certain  limits" 
(Smith).  In  order  to  understand  more  clearly  the  technical 
terms  applied  to  solutions  and  also  their  meaning,  the  follow- 
ing definitions  are  given.  It  is  customary  to  speak  of  a 


SOLUTIONS  3 

substance  which,  like  water,  forms  the  bulk  of  the  solution  as 
the  solvent.  The  substance  which  is  dissolved  is  called  the 
solute.  The  amount  of  substance  which  is  dissolved  in  a  cer- 
tain quantity  of  the  solvent  may  vary  between  wide  limits  with 
some  substances,  while  with  others  it  is  more  limited.  A 
solution  which  contains  a  small  portion  of  the  dissolved  body 
is  spoken  of  as  dilute.  A  solution  that  contains  large  amounts 
of  the  dissolved  substance  is  called  a  concentrated  solution. 
The  amount  of  material  contained  in  a  given  volume  of 
solution  is  spoken  of  as  the  strength  of  that  solution.  The 
terms  dilute  and  concentrated  do  not  tell  exactly  the  amount 
of  substance  in  solution.  For  more  exact  work  it  is  necessary 
to  use  solutions  the  exact  strength  of  which  can  be  defined. 
A  convenient  standard  frequently  taken  is  one  gram-molecule 
of  the"  substance  dissolved  in  enough  water  to  make  the  total 
volume  one  liter.  This  solution  would  be  of  one  molar  con- 
centration. A  solution  that  contained  one  tenth  of  a  molec- 
ular weight  in  enough  water  to  make  the  total  volume  one 
liter  would  be  one-tenth  molar. 

The  freezing  point  of  pure  water  is  0°  centigrade,  and  its 
boiling  point  at  760  mm.  pressure  is  100°  centigrade.  If 
some  nonvolatile  substance  is  dissolved  in  pure  water,  the 
freezing  point  of  the  solution  is  found  to  be  below  0°  and 
its  boiling  point  is  above  100°,  the  change  being  proportional 
to  the  amount  of  solute  in  a  given  volume.  It  was  shown 
by  Raoult  in  1886  that  in  dilute  solutions  of  different  sub- 
stances, of  the  same  molar  strength,  the  boiling  points  of  the 
different  solutions  would  be  raised  the  same  amount  for  each 
substance.  Molar  solutions  of  cane  sugar  (C12H22On  =  342  g.) 
or  glycerin  (C3HgO3  =  92  g.)  in  water  will  each  have  a  boil- 
ing point  of  100.52°.  If  the  temperature  at  which  these  two 
solutions  freeze  is  measured,  it  is  found  to  be  —  1.86°  in  each 
case.  If  the  solutions  are  half  molar,  the  boiling  point  will  be 
100.26°  and  the  freezing  point  —  0.93°.  Other  measurements 


4  NOTES  ON  QUALITATIVE  ANALYSIS 

on  solutions  of  various  strengths  will  show  the  elevation  of 
the  boiling  point  and  the  depression  of  the  freezing  point  to 
be  proportional  to  the  molar  strength  of  the  solutions.  Solu- 
tions of  the  same  molar  strength  of  all  substances  that  are 
nonvolatile  and  whose  solutions  do  not  conduct  electricity 
behave  as  glycerin  and  cane  sugar. 

Avogadro's  hypothesis  states  that  equal  volumes  of  all 
gases  under  the  same  conditions  of  temperature  and  pres- 
sure contain  the  same  number  of  molecules.  As  the  gram- 
molecular  weight  of  any  gas  always  occupies  the  same  volume 
(22.4  liters),  it  follows  that  all  gram-molecular  weights -con- 
tain the  same  number  of  molecules ;  hence  all  molar  solutions 
contain  the  same  number  of  particles  in  equal  volumes  of 
solution.  The  change  in  freezing  point  and  boiling  point  is, 
therefore,  proportional  to  the  number  of  molecules,  or  parti- 
cles of  solute  in  a  given  volume  of  solution. 

Experimental  results  show  that  some  molar  solutions  give 
abnormally  large  changes  of  freezing  and  boiling  points,  and 
that  these  are  the  only  solutions  which  conduct  electricity. 
If  an  electric  current  is  passed  through  such  a  solution  it  is 
found  that  the  metallic  part  of  the  molecule  will  collect 
around  the  negative  electrode  (cathode),  and  the  nonmetallic 
part  will  collect  around  the  positive  electrode  (anode).  This, 
would  lead  us  to  think  that  the  molecule  is  in  some  manner 
dissociated  into  parts  in  the  solution.  It  is  believed  that  one 
of  these  divisions  of  the  molecule  carries  a  positive  electric 
charge,  and  the  other  a  negative  electric  charge,  or  they 
would  not  be  attracted  to  the  oppositely  charged  electrodes. 
These  charged  divisions  of  the  molecule  are  called  ions. 
This  dissociation  in  solution  is  shown  almost  exclusively  by 
acids,  bases,  and  salts,  and  such  substances  are  called  iono- 
gens.  The  molecule  of  sodium  chloride  on  this  basis  would 
divide  into  two  parts,  NaCl  "< — *  Na+  +  Cl~.  If  all  of  the 
molecules  are  dissociated  at  the  same  time,  a  solution  would 


DISSOCIATION 


contain  twice  as  many  ions  as  the  number  of  molecules  of 
undissociated  salt.  Since  the  freezing-point  lowering  and 
the  boiling-point  elevation  are  proportional  to  the  concentra- 
tion of  particles  in  solution,  a  solution  of  sodium  chloride 
would  give  twice  the  freezing-point  depression  and  boiling- 
point  elevation  that  solutions  of  the  same  molar  strength  of 
cane  sugar  show.  Experiments  show  that  in  very  dilute  solu- 
tions this  is  approximately  true.  Magnesium  chloride  shows 
the  possibility  of  dissociating  into  three  parts  (MgCl2  ^  *"  Mg++ 
+  Cl~  +  Cl~).  Very  dilute  solutions  show  approximately  three 
times  the  normal  value.  These  values,  however,  are  never 
quite  twice  or  three  times  the  normal  but  approach  these 
limits  as  the  solutions  become  more  dilute.  Hence  we  assume 
that  only  a  part  of  the  molecules  are  broken  up  at  any  one 
time.  The  following  table  shows  approximately  the  percentage 
of  molecules  ionized  in  a  few  solutions  of  various  strengths : 


DILUTION 

HC1 

H2S04 

HC2H302 

NaCl 

.IN    

92 

61 

1.3 

84 

.01  N       .... 

96 

83 

4.1 

93 

.001  N     .... 

100 

98 

11.6 

98 

The  fact  of  evaporation  indicates  that  the  molecules  of  a 
liquid  as  well  as  the  molecules  of  a  gas  are  in  motion.  If 
we  assume  that  the  molecules  of  a  Liquid  are  in  motion,  then 
we  have  a  plausible  explanation  for  the  evaporation  of  a 
liquid  as  well  as  other  phenomena  shown  by  liquids.  In  the 
collisions  and  jostlings  of  the  moving  molecules  some  will 
acquire  velocities  greater  than  the  average  velocity.  Some 
of  these  molecules,  because  of  their  unusual  velocity,  will 
break  away  from  the  attraction  of  their  neighbors  and  fly  off 
into  the  space  above  the  liquid.  If  a  current  of  rapidly 
moving  air  is  blown  across  the  surface  of  the  liquid,  the 


6  NOTES  ON  QUALITATIVE  ANALYSIS 

escaping  molecules  will  be  carried  away,  causing  evaporation 
to  be  more  rapid.  If  the  liquid  is  placed  in  a  closed  vessel 
the  molecules  cannot  escape  from  the  inclosed  space,  and 
some  of  them,  by  collision  with  other  molecules  and  the 
walls  of  the  vessel,,  will  be  deflected  back  to  the  surface  of 
the  liquid  and  condensed.  As  the  concentration  of  molecules 
above  the  liquid  increases,  the  number  of  these  collisions  will 
increase,  thus  causing  the  number  of  molecules  returning  to 
the  liquid  in  a  given  time  to  increase.  A  condition  must 
finally  result  in  which  the  number  returning  to  the  liquid  in 
a  given  time  is  just  equal  to  the  number  escaping  from  the 
liquid  in  the  same  length  of  time.  When  this  condition 
exists  and  equilibrium  is  established,  and  the  space  above 
the  liquid  said  to  be  saturated  with  the  liquid  vapor,  the 
vapor  phase  must  remain  constant,  because  the  number  of 
molecules  added  to  the  vapor  in  a  given  time  is  equal  to  the 
number  taken  away  in  the  same  length  of  time.  Evaporation 
has  not  ceased,  but  evaporation  and  condensation  are  going  on 
at  the  same  rate.  If  the  space  above  the  liquid  is  suddenly 
increased  the  equilibrium  is  destroyed  until  enough  molecules 
have  vaporized  to  bring  the  concentration  of  the  vapor  again 
to  the  saturated  condition.  If  the  space  is  suddenly  decreased, 
equilibrium  is  destroyed  and  will  be  reestablished  only  when 
enough  molecules  have  condensed  to  bring  the  concentration 
of  the  vapor  again  to  the  saturated  condition.  When  two 
opposing  reactions  are  in  equilibrium,  so  that  any  change  in 
conditions  tends  to  favor  one  reaction  or  the  other,  such  a 
reaction  is  said  to  be  reversible.  Thus  the  evaporation  of  water 
is  reversible,  and  may  be  represented  by  the  equation, 

H2O  (liquid)  :<=>:  H2Q  (vapor). 

Similarly  to  the  evaporation  of  a  liquid,  the  process  of  a 
solid  dissolving  in  a  liquid  may  be  considered  to  be  a  dif- 
fusion of  the  molecules  of  the  solid  among  the  molecules  of 


EQUILIBRIUM  7 

the  liquid.  Since  numerous  collisions  will  take  place,  some 
of  the  molecules  will  be  deflected  in  such  a  manner  as  to 
return  to  the  solid,  while  others  will  be  diffused  through  the 
liquid  in  all  directions.  With  most  substances  it  is  possible 
to  increase  the  concentration  of  the  molecules  in  solution  to 
such  an  extent  that  the  number  returning  to  the  solid  in  a 
given  time  will  just  equal  the  number  diffusing  away  from 
the  solid  in  the  same  time.  Under  such  a  condition  equilib- 
rium will  exist  between  the  solid  and  its  solution : 

Solid  <    >  solution. 

If  a  nonvolatile  substance,  such  as  sodium  chloride,  is 
dissolved  in  water,  and  the  water  removed-  gradually  by 
evaporation,  the  concentration  of  the  salt  in  solution  must 
increase.  When  the  concentration  has  reached  a  certain 
value  some  of  the  salt  will  appear  in  the  solid  form.  This 
means  that  the  concentration  of  the  dissolved  salt  is  so  great 
that  the  number  of  molecules  that  tend  to  go  to  the  solid 
form  in  a  given  time  is  greater  than  the  number  that  would 
be  dissolved  in  the  same  time,  thus  resulting  in  the  removal 
of  some  of  the  dissolved  salt  by  the  formation  of  the  solid. 
If  water  is  added  to  the  solution,  equilibrium  will  be  dis- 
placed in  the  opposite  direction  and  the  solid  will  go  into 
solution.  This  reaction  is  then  reversible: 

NaCl  (solid)  :<=>:  NaCl  (dissolved). 

Some  of  our  most  familiar  chemical  reactions  are  easily 
reversible.  If  we  pass  steam  through  a  tube  containing 
heated  iron  filings  the  following  reaction  occurs: 

3  Fe  +  4  H2O >-  Fe8O4  +  4  H2 ; 

and  the  hydrogen  may  be  collected  over  water.  If  we  pass 
hydrogen  over  heated  iron  oxide  this  reaction  takes  place: 

Fe804  +  4  H2  — >•  3  Fe  +  4  H2O. 


8  NOTES  ON  QUALITATIVE  ANALYSIS 

These  reactions  are  the  reverse  of  each  other  and  apparently 
are  contradictory  it  will  be  observed. 

Why  will  not  the  hydrogen  and  iron  oxide  formed  in  the 
first  equation  react  and  reverse  that  equation  ?  The  answer 
is  that  they  do.  If  hydrogen  and  iron  oxide  are  sealed  up 
in  a  tube  and  heated,  we  find  that  iron  and  water  vapor  are 
formed  but  that  there  is  also  some  iron  oxide  and  hydrogen 
left.  Similarly,  if  iron  and  water  vapor  are  heated  in  a  sealed 
tube,  hydrogen  and  iron  oxide  are  formed,  but  much  of  the 
iron  and  water  vapor  exists  uncombined.  Indeed,  if  the  two 
tubes  are  heated  to  the  same  temperature  under  the  same 
conditions,  the  weight  of  each  product  in  one  tube  would  be 
the  same  as  the  weight  of  the  corresponding  product  in  the 
other  tube. 

The  explanation  is  that  in  one  tube  the  hydrogen  and  iron 
oxide  begin  to  react  to  form  iron  and  water  vapor,  but  that 
as  soon  as  these  substances  are  formed  they  in  turn  react 
with  each  other  to  form  hydrogen  and  iron  oxide  again.  In 
the  other  tube  the  hydrogen  and  iron  oxide  formed  react  to 
form  iron  and  water  vapor,  so  that  in  each  tube  both  reac- 
tions are  going  on  at  the  same  time.  Just  as  in  the  evap- 
oration and  condensation  of  water,  there  must  be  a  point  at 
which  these  two  opposing  reactions  exactly  balance  each 
other.  When  steam  is  passed  over  iron  for  the  purpose  of 
obtaining  hydrogen,  the  large  excess  of  steam  used  carries 
along  with  it  the  comparatively  small  amount  of  hydrogen 
formed,  and  the  reverse  reaction  is  not  noticeable.  In  passing 
hydrogen  over  iron  oxide  there  is  a  large  excess  of  hydrogen 
compared  with  the  amount  of  steam  formed,  and  the  reverse 
action  is  slight.  It  is  seen  that  the  mass  of  the  steam  or 
hydrogen  determines  the  direction  of  the  reaction. 

When  hydrogen  chloride  is  prepared  by  the  reaction, 

NaCl  +  H2SO4  — *  NaHSO4  +  HC1, 


THE  MASS  LAW  9 

the  action  apparently  goes  to  completion,  and  the  final  prod- 
ucts are  sodium  hydrogen  sulfate  and  hydrogen  chloride. 
If,  however,  hydrogen  chloride  is  added  to  a  strong  solution 
of  sodium  hydrogen  sulfate  a  copious  precipitate  of  sodium 
chloride  is  produced,  showing  that  the  reaction, 


NaHS04  +  HC1  -  ^NaCl  +  H2SO4, 

has  taken  place.  The  first  reaction  must  therefore  be  reversi- 
ble, but  under  the  conditions  existing  when  hydrogen  chloride 
is  prepared,  the  speed  in  one  direction  must  be  far  greater 
than  in  the  other.  This  is  easily  explained  when  it  is  remem- 
bered that  hydrogen  chloride  is  a  gas,  and  at  the  tempera- 
ture at  which  the  reaction  is  usually  carried  out  is  practically 
insoluble,  hence  none  is  left  in  contact  with  the  sodium 
hydrogen  sulfate  to  reverse  the  reaction. 

From  these  examples  it  is  easily  seen  that  the  direction  of 
a  reversible  reaction  depends  upon  the  concentrations  of  the 
reacting  substances.  This  is  known  as  the  Mass  Law  of 
Guldberg  and  Waage,  which  they  stated  as  follows:  The 
speed  of  a  reaction  is  proportional  to  the  concentrations  of  the 
reacting  substances. 

By  a  study  of  the  table  of  ionization  with  dilution  given 
on  page  5,  it  will  be  seen  that  ionization  varies  with  dilution. 
With  greater  dilution  more  of  the  molecules  are  constantly 
in  the  form  of  ions,  while  with  more  concentrated  solutions 
the  amount  of  ionization  is  less.  According  to  the  idea  of 
reversibility  the  reaction  qf  an  ionogen  dissociating  into  its 
ions  is  reversible.  For  example,  sodium  chloride  dissociates 
into  its  ions,  and  the  ions  recombine  with  a  certain  velocity 
depending  on  their  concentration  according  to  the  equation, 

NaCl  :«z±  Na+  +  Cl~. 

The  following  are  some  typical  equations  showing  the  ioniza- 
tion in  solution  of  a  few  familiar  acids,  bases,  and  salts  : 


10  NOTES  ON  QUALITATIVE  ANALYSIS 

HC1  +=±  H+  +  C1-. 
NH4OH  +=±  NH4+  +  OH-. 

FeCl8  ^=t  Fe+++  +  Cl-  +  Cl~ 
MgCl2  +=±  Mg++  +  Cl-  4-  C1-. 
MgS04  +=+  Mg++  +  S04—  . 


There  is  great  variation  shown  in  the  degree  to  which 
these  substances  are  ionized  in  solutions  of  the  same  molar 
concentration. 

When  equilibrium  is  established  between  hydrogen  chloride 
and  its  ions  in  a  .1  molar  solution,  about  90  per  cent  of  the 
molecules  are  dissociated  into  ions.  In  a  .1  molar  solution 
of  acetic  acid  about  1.3  per  cent  is  dissociated  into  the  form 
of  ions.  A  study  of  the  properties  of  these  acids  shows  that 
hydrochloric  acid  is  a  strong  acid,  while  acetic  acid  is  a  weak 
acid.  It  is  found  in  every  instance  that  strong  acids  are 
those  that  are  highly  ionized,  while  the  weak  acids  are  those 
that  are  poorly  ionized.  Similarly,  the  strong  bases  are  the 
bases  that  are  highly  ionized,  while  the  weak  bases  are  those 
that  are  poorly  ionized.  Very  nearly  all  salts  are  well  ionized 
and  to  about  the  same  degree.  The  following  table  gives  the 
approximate  ionization  of  various  acids  and  bases  in  .1  molar 
solutions  : 

PERCENTAGE  PERCENTAGE 

IONIZED  IONIZED 

NaOH     ........     90  H(SO4-)      .......     61 

KOH  .........     90  HC2H8O2    .......    1.3 

NH4OH       .......    1.3  H(HS)    .......  0.1-0.2 

HC1    .........     90  H(HCO3)    ......  0.1-0.2 

HNO8     ........     90  HCO8-    ........  005 

H(HSO4)    .......     90  HS-   ........  0.0001 

From  the  foregoing  discussion  it  is  evident  that  when  an 
ionogen  is  dissolved  in  water  the  solution  does  not  consist 
merely  of  the  molecules  of  the  substance  in  water,  but  is 


REVERSIBLE  REACTIONS  11 

really  more  complex,  containing  also  the  ions  into  which  the 
salt  dissociates.  Thus  a  solution  of  sodium  chloride  will 
contain  NaCl  molecules,  Na+  ions,  and  .Cl~  ions  ;  a  solution 
of  potassium  nitrate  KNO3  molecules,  K+  ions,  and  NO8~  ions  : 
a  solution  of  sulfuric  acid  H2SO4  molecules,  H+  ions,  HSO4~ 
ions,  and  SO4  ions.  Frequently  we  have  to  take  into  con- 
sideration also  the  fact  that  water  is  very  slightly  dissociated 
into  H+  ions  and  OH~  ions. 

In  reactions  between  ionogens  in  solution  all  of  these  sub- 
stances must  be  taken  into  consideration. 

When  chemical  reactions  take  place  between  ionogens  in 
solution  the  reaction  is  generally  a  double  decomposition 
reaction.  Reactions  of  this  type  always  take  place  by  the 
dissociation  of  the  molecules  into  ions,  and  the  ions  recom- 
bining  in  other  combinations  forming  other  molecules  than 
the  original.  If  solutions  of  sodium  chloride  and  potassium 
nitrate  are  mixed  the  reactions  will  take  place  according  to 
the  following  equation: 

NaCl  ^=t   Na+    -f  Cl~ 
KNO,  +=±  NO  -  +  K+ 

ti   ii 

NaN03  +  KC1. 

It  follows  from  the  Mass  Law  that  the  proportion  of  each  sub- 
stance in  the  solution  after  mixing  depends  upon  the  concen- 
trations of  the  various  ions.  Concentration  means  the  amount 
in  unit  volume  without  regard  to  the  total  quantity  present. 
If  ferric  chloride  and  potassium  thiocyanate  solutions  are 
mixed,  a  deep  blood-red  compound  of  ferric  thiocyanate  is 
formed  : 


3  KCL 

This  reaction  is  reversible.  If  to  the  solution  some  solid 
KC1  is  added,  it  will  dissolve,  increasing  the  concentration  of 
KC1  in  the  solution,  which  will  cause  the  reverse  reaction 


12  NOTES  ON  QUALITATI\7E  ANALYSIS 

to  take  place.  This  will  decrease  the  red  Fe  (CNS)3,  and 
consequently  the  color  of  the  solution  will  become  lighter.  If, 
on  the  other  hand,  the  concentration  of  either  FeCl3  or  KCNS 
is  increased  by  adding  more  of  either  substance,  the  amount 
of  Fe  (CNS)8  increases  and  the  color  deepens. 

In  any  reversible  reaction  a  definite  relation  is  found  to 
exist  between  the  concentrations  of  the  reacting  substances 
and  the  products  of  the  reaction.  This  can  be  stated  in  the 
following  mathematical  terms.  The  product  of  the  concen- 
trations of  the  reacting  substances  divided  by  the  product 
of  the  concentrations  of  the  resulting  substances  is  equal 
to  a  constant  quantity.  If  any  concentrations  are  changed, 
the  other  concentrations  will  change  in  such  a  manner  that 
the  above  condition  will  still  hold  true.  The  condition  for 
equilibrium  in  the  reaction  of  ferric  chloride  and  potassium 
thiocyanate  is  expressed  by  the  equation, 


KCN8 


^  KC1  A 

where  the  factors  C  represent  the  concentrations  of  the  re- 
acting substances.  The  concentration  must  be  raised  to  a 
power  equal  to  the  number  of  molecules  of  the  substance 
considered.  The  factor  K  will  remain  constant  for  different 
concentrations  of  the  reacting  components.  With  sodium 
chloride  in  aqueous  solution  the  chemical  equation  would  be 
NaCl  <  >  Na+  H-  Cl~,  and  the  mathematical  equation  for  a 
condition  of  equilibrium  would  be 

x  Ccl-     ~ 
~  --       ' 


where  CNa+  represents  the  concentration  of  the  sodium  ion, 
Ca-  the  concentration  of  the  chlorine  ion,  and  CNaC1  the  con- 
centration of  the  non-ionized  molecules.  The  constant  K  is 
called  the  ionization  or  dissociation  constant.  For  a  highly 


EQUILIBRIUM  13 

dissociated  substance,  as  sodium  chloride,  the  constant  varies 
considerably  with  change  of  concentration,  but  holds  very 
closely  for  little-ionized  substances. 

If  3.5  grams  of  ammonium  hydroxide  is  dissolved  in  enough 
water  to  make  a  liter  of  solution,  experiments  show  when 
equilibrium  is  established  that  1.31  per  cent  of  the  3.5  grams 
is  in  the  form  of  ions,  while  98.69  per  cent  is  in  the  form  of 
non-ionized  molecules : 

NH4OH  +=>:  NH4+  +  OH". 

1.31%  1.31%  ' 


As  the  gram-molecular  weight  of  ammonium  hydroxide  is 
35,  this  solution  is  .1  molar.  The  molar  concentration  of  the 
ions  will  be  (0.1  x  .0131)  =  0.00131  molar;  the  molar  con- 
centration of  the  non-ionized  molecules  will  be  (0.1  x  .9869) 
=  0.09869  molar.  Substituting  these  values  in  the  equation, 


Thus  the  value  of  K  is  found  to  be  0.000017+. 

In  a  solution  containing  0.35  grams  of  ammonium  hydroxide 
per  liter  (0,01  molar),  4.07  per  cent  is  in  the  form  of  ions 
and  95.93  per  cent  is  in  the  form  of  non-ionized  molecules. 
Substituting  these  values  in  the  equation, 

(0.01  x  0.0407)  x  (0.01  x  0.0407)  =  K  =  0  000017+. 

thus  the  value  of  K  is  found  to  be  the  same  as  for  a  0.1  molar 
solution.  If  many  solutions  of  varying  concentrations  of 
ammonium  hydroxide  are  measured,  the  values  of  K  obtained 
from  all  will  be  found,  within  the  limits  of  ordinary  experi- 
mental work,  to  be  0.000018.  As  already  stated,  this  rule 
holds  for  all  slightly  ionized  or  difficultly  soluble  ionogens 


14  NOTES  ON  QUALITATIVE  ANALYSIS 

in  aqueous  solution,  each  ionogen  showing  its  own  particular 
constant  which  is  different  from  all  others. 

The  concentration  in  the  dissociation-constant  equation 
represents  the  total  concentration  of  that  particular  ion  or 
molecule,  regardless  of  whether  it  comes  from  one  or  more 
substances  that  are  in  solution.  Thus  in  a  solution  contain- 
ing NH4OH  and  NH4C1  the  value  of  the  concentration  of 
the  NH4+  ion  is  made  up  of  the  concentration  of  the  NH4+ 
ion  from  NH4OH  plus  the  concentration  of  the  NH4+  ion 
from  NH4C1. 

If  to  a  0.1  molar  solution  of  ammonium  hydroxide  some 
strongly  ionized  ammonium  chloride  is  added,  the  concentra- 
tion of  the  NH4+  ion  is  greatly  increased  and  K  becomes 
greater  than  0.  00001  7+.  Equilibrium  can  only  be  established 
when  the  OH~  ions  have  so  far  decreased  by  combining  with 
NH4+  ions  that  K  again  equals  0.00001  7+.  As  the  basic  prop- 
erties of  ammonium  hydroxide,  or  any  base,  are  due  to  the 
OH~  ions  which  it  contains,  the  addition  of  NH4+  ions  (from 
NH4C1  or  any  other  ammonium  salt)  makes  ammonium  hy- 
droxide appear  as  a  weaker  base  because  of  the  corresponding 
decrease  in  OH~  ions. 

Similarly,  acetic  acid  (H  -  C2H8O2),  which  ionizes 


may  be  made  to  appear  as  a  weaker  acid  by  adding  sodium 
acetate  (Na  •  C2H8O2),  which  increases  the  C2H3O2~  ions  in 
solution  and  correspondingly  decreases  the  H+  ion  charac- 
teristic of  acids.  The  concentration  of  any  ion  in  a  solution, 
then,  may  be  decreased  by  adding  an  excess  of  the  other  ion 
into  which  the  substance  dissociates.  Similarly,  the  concen- 
tration of  any  ion  in  a  solution  may  be  increased  by  remov- 
ing the  other  ion  into  which  the  substance  dissociates.  Ions 
may  be  removed  from  a  solution  by  causing  them  to  combine 
to  form  insoluble  or  little-ionized  substances. 


EQUILIBRIUM  15 

The  rule  for  bringing  reactions  to  completion  by  form- 
ing insoluble,  volatile,  or  little-ionized  substances  depends 
on  this  fact. 

The  direction  in  which  an  ionic  reaction  will  proceed  can 
be  predicted  if  the  degree  of  ionization  and  the  solubility  of 
the  reaction  compounds  are  known.  In  the  reaction  of  sodium 
hydroxide  and  hydrochloric  acid  the  reaction  will  go  prac- 
tically to  completion  in  the  direction  which  results  in  the 
formation  of  water  —  a  very  little-ionized  substance: 

Na+  +  OH-  +  H f  +  Cl-  +=±  H2O  +  Na+  +  Cl~. 
The  sodium  hydroxide  and  hydrochloric  acid  are  highly 
ionized,  and  the  ions  formed  recombine  in  another  combina- 
tion to  form  water  and  sodium  chloride.  If  water  ionizes 
into  hydroxyf  ions  and  hydrogen  ions,  and  sodium  chloride 
is  also  ionized,  there  would  be  no  reason  to  think  that  these 
ions  would  not  recombine  to  form  sodium  hydroxide  and 
hydrochloric  acid,  thus  making  the  reaction  reversible.  How- 
ever, water  gives  exceedingly  few  ions ;  hence  the  formation 
of  sodium  hydroxide  and  hydrochloric  acid  is  possible  only 
to  a  very  slight  extent.  The  reaction  must  then  go  practically 
to  completion  in  the  direction  which  results  in  the  formation 
of  water.  In  the  reaction  of  silver  nitrate  and  hydrochloric 
acid  the  reaction  will  go  practically  to  completion  with  the 
formation  of  silver  chloride  and  nitric  acid : 

Ag+  +  N03-  +  H+  +  Cl-  =<=>  AgCl  +  H+  +  NO,-. 
The  reason  why  this  reaction  goes  to  completion  is  that 
silver  chloride  is  insoluble.  The  silver  ions  and  the  chlorine 
ions  will  be  removed  from  solution  in  the  formation  of  the 
insoluble  silver  chloride,  and  therefore  cannot  take  part  in  the 
reverse  reaction.  In  the  reaction  of  hydrochloric  acid  on  cal- 
cium carbonate  the  reaction  will  go  to  completion  with  the 
formation  of  calcium  chloride,  water,  and  carbon  dioxide: 

2  HC1  +  CaC03  =  CaCL  +  H.CO.  (H2O  +  COa). 


16  NOTES  ON  QUALITATIVE  ANALYSIS 

Carbonic  acid  is  formed,  which  decomposes  into  carbon  diox- 
ide and  water,  thus  removing  the  carbonate  ions  and  hydrogen 
ions  from  solution.  In  these  three  illustrations  it  is  shown 
that  the  condition  under  which  an  ionic  reaction  goes  to  com- 
pletion is  the  formation  of  some  compound  that  for  one  or 
more  of  three  reasons  does  not  give  ions  in  quantity  enough 
to  cause  the  reaction  to  go  in  the  reverse  direction.  These 
three  conditions  are  (1)  the  formation  of  a  substance  that 
does  not  ionize,  (2)  the  formation  of  a  substance  that  is 
insoluble,  (3)  the  formation  of  a  gas.  In  none  of  these  cases 
can  the  reaction  be  said  to  go  absolutely  to  completion,  be- 
cause water  is  slightly  ionized,  silver  chloride  is  very  slightly 
soluble,  and  carbonic  acid  is  not  completely  decomposed  at 
ordinary  temperature.  Reactions  of  this  type  #re  said  to  go 
to  completion  because  the  reverse  action  is  so  slight  that  it 
•  does  not  interfere  in  ordinary  work. 

If  a  solution  of  sodium  acetate  is  treated  with  hydro- 
chloric acid,  the  reaction  will  reach  equilibrium  when  very 
nearly  all  of  the  acetate  ions  are  combined  with  hydrogen 
ions  and  very  few  are  free  in  the  solution : 

HC1  +  NaC2H8O2  q=±  NaCl  +  HC2H8O2. 

The  reason  for  this  is  because  acetic  acid  is  the  least  ionized 
of  any  of  the  reacting  substances  present,  as  shown  by  the 
following  ionization  constants: 

H+xCl-  Na+xCHO- 

=  greater  than  1.    •  A  A       =  approximately  .5. 


NaCl    =  Beater  than  1.     "  "  -  0.000018. 


In  a  reaction  where  a  strong  acid  acts  on  the  salt  of  a  weaker 
acid,  the  salt  of  the  weak  acid  will  be  decomposed  when  the 
weak  acid  gives  fewer  ions  to  the  solution  than  its  salt. 


SOLUBILITY  PRODUCT  17 

SOLUBILITY-PRODUCT  PRINCIPLE 

When  water  is  added  to  solid  calcium  sulfate  the  salt  will 
dissolve  to  a  limited  extent.  If  an  excess  of  solid  salt  is 
used,  equilibrium  will  result  between  the  solid  salt  and  its 
solution  : 


The  dissolved  salt  is  in  equilibrium  with  its  ions: 
CaSO4  (dissolved)  T~*"  Ca++  +  SO4~. 
The  complete  equation  is 

CaSO4  (solid)  -<~>  CaSO4  (dissolved)  -<—*•  Ca++  +  SO4~. 
The  ionization  constant  equation  for  calcium  sulfate  is 

Ca++xS04-- 
CaSO4 

In  a  saturated  solution  the  concentration  of  the  solute  re- 
mains constant  at  a  given  temperature.  The  concentration 
of  undissociated  molecules  also  remains  constant.  The  above 
equation  may  then  be  written  for  a  saturated,  solution 

Ca++xSO4-       Ca++xSO-        v  Jjr     v 

CaS044     =  -  ir^-=^orC^xSO-  =  K1xKJ. 

Since  the  product  of  two  constants  is  a  constant,  then 
Ca++  x  S04—  =  K. 

Stated  in  words,  in  saturated  solutions  of  difficultly  soluble 
salts  the  product  of  the  concentration  of  the  ions  equals  a  con- 
stant value.  This  is  the  solubility-product  principle.  A  satu- 
rated solution  of  strontium  sulfate  contains  0.00062  moles  per 
liter,  85  per  -cent  of  which  is  ionized.  The  concentration  of 
Sr++  =  SO4—  would  be  0.00062  x  0.85  =  0.0005. 

Sr++  x  S.O4—  =  (0.0005)2  =  0.00000025. 

If  the  product  of  strontium  ions  and  sulfate  ions  in  solution 
exceeds  this  value,  the  ions  in  excess  will  combine  to  form 


18  NOTES  ON  QUALITATIVE  ANALYSIS 

undissociated  molecules,  and  if  the  solution  is  saturated,  some 
of  the  undissociated  molecules  must  separate  out  of  solution 
as  solid  strontium  sulfate.  Whenever  the  product  of  the 
concentration  of  the  ions  exceeds  the  value  of  the  solubility 
product  constant  precipitation  will  occur  unless  a  super- 
saturated solution  is  formed.  The  importance  of  these  prin- 
ciples,' and  their  application  to  analysis,  will  become  evident 
as  we  proceed  and  should  be  continually  kept  in  mind. 

HYDROLYSIS 

In  studying  hydrogen  iodide  it  was  pointed  out  that  phos- 
phorous iodide  and  water  react  to  form  hydrogen  iodide  and 
phosphorous  acid : 

PI8  +  3  H2O  =  3  HI  +  P(OH)8. 

This  type  of  reaction  was  called  hydrolysis. 

If  normal  sodium  carbonate  (Na2COg)  is  dissolved  in  dis- 
tilled water,  the  solution  will  react  alkaline  to  litmus,  but 
if  normal  copper  sulfate  (CuSO4)  is  dissolved  in  distilled 
water,  the  solution  reacts  acid  to  litmus.  These  facts  are 
easily  understood  when  it  is  remembered  that  water  is  very 
slightly  dissociated  into  hydrogen  and  hydroxyl  ions : 

H2O  +=±  H+  x  OH-. 

The  amount  of  this  dissociation  is  approximately  one  gram 
molecule  in  ten  million  liters  of  water.  In  consequence  of 
this  ionization,  although  it  is  extremely  small,  water  will 
enter  into  double  decomposition  reactions  with  other  ionogens. 
It  follows  that  when  normal  sodium  carbonate  is  dissolved  in 
water  a  reaction  will  take  place, 

Na2C08  +  2  HOH  ^=±:  2  NaOH  +  H2CO3, 

sodium  hydroxide  and  carbonic  acid  being  formed  in  equal 
amounts.  Although  water  is  very  little  ionized,  and  the 


HYDROLYSIS  19 

reaction  proceeds  very  slightly  toward  the  right,  some  sodium 
hydroxide  and  carbonic  acid  are  formed.  Sodium  hydroxide 
is  a  strong  base  and  gives  many  hydroxyl  ions,  while  car- 
bonic acid,  being  a  very  weak  acid,  gives  few  hydrogen  ions. 
It  is  evident  there  are  more  hydroxyl  ions  in  the  solution 
than  there  are  hydrogen  ions,  which  will  cause  the  solu- 
tion to  react  alkaline.  For  the  same  reason,  sodium  acetate 
(NaC2H3O2),  calcium  sulfide  (CaS),  and  many  other  salts 
show  alkaline  reaction  when  dissolved  in  water.  Copper  sul- 
fate  when  dissolved  in  water  reacts  according  to  the  following 
equation, 

CuS04  +  2  HOH  +=±  Cu  (OH)2  +  H2SO4. 

As  in  the  case  of  sodium  carbonate  and  water,  the  reaction 
proceeds  very  slightly  to  the  right,  but  equal  amounts  of 
copper  hydroxide  and  sulfuric  acid  are  formed.  Copper 
hydroxide  is  almost  insoluble  and  a  weak  base  and  so  gives 
few  hydroxyl  ions,  while  sulfuric  acid  is  easily  soluble  and  a 
very  strong  acid,  giving  many  hydrogen  ions.  The  excess  of 
hydrogen  ions  over  the  hydroxyl  ions  causes  the  solution  to 
react  acid  to  litmus.  Other  salts  which  react  acid  for  the 
same  reason  when  dissolved  in  water  are  aluminium  chloride 
(A1C13),  zinc  sulfate  (ZnSO4),  mercury  nitrate  (Hg(NO3)2), 
etc.  If  the  salt  that  is  dissolved  in  water  is  formed  from  an 
acid  and  a  base,  both  of  which  are  highly  ionized,  hydrolysis 
will  not  take  place  to  the  extent  it  would  if  the  acid  or  base, 
or  both,  were  slightly  ionized,  because  the  concentration  of 
hydrogen  and  hydroxyl  ions  from  the  strongly  ionized  acid 
and  base  tends  to  depress  the  ionization  of  water ;  also,  as 
both  acid  and  base  are  ionized  to  about  the  same  extent,  the 
solution  will  react  neutral.  Sodium  chloride  may  be  taken 
as  an  example  of  this  class.  If  both  the  acid  and  the  base 
that  form  the  salt  are  slightly  ionized,  hydrolysis  will  be 
marked. 


20  NOTES  ON  QUALITATIVE  ANALYSIS 

OXIDATION  AND  REDUCTION 

The  definitions  of  oxidation  as  the  addition  of  oxygen  to  an 
element  or  compound,  and  of  reduction  as  the  removal  of  oxy- 
gen from  a  compound,  express  incompletely  what  is  meant  by 
these  terms.  Oxidation,  more  broadly  speaking,  is  an  increase 
in  the  positive  valence  of  an  element  or  radical,  while  reduction 
is  a  decrease  in  the  positive  valence  or  an  increase  in  the  nega- 
tive valence.  If  chlorine  is  added  to  a  solution  of  ferrous  chlo- 
ride, a  reaction  takes  place  in  which  the  ferrous  iron  is  changed 
to  ferric  iron,  g  ^  +  clj  _^  2  FeCla ; 

or,  written  from  the  ionic  viewpoint, 

2  Fe++  +  4  Cl-  +  C12 *•  2  Fe+++  +  6  Cl~. 

The  two  ferrous  ions  have  each  gained  a  positive  charge, 
while  the  chlorine  molecule  has  now  become  two  negatively 
charged  ions.  The  positive  valence  of  the  iron  ion  has  thus 
increased  from  two  to  three.  This  is  oxidation  of  ferrous  iron. 

If  ferric  chloride  is  treated  with  hydrogen  sulfide  in  acid 
solution,  the  following  reaction  takes  place: 

2  FeCl,  +  H2S +  2  FeCl2  +  2  HCl-f  S ; 

or,  written  from  the  ionic  viewpoint, 
2  Fe++++6  C1-+  2  H++S-    ->2  Fe+++ 6  C1-+2  H++S. 

A  positive  charge  from  each  ferric  ion,  and  the  two  negative 
charges  of  the  sulfur  ion,  have  neutralized  each  other.  The 
iron  has  thus  had  its  positive  valence  reduced  from  three  to 
two  and  the  sulfur  has  been  changed  from  a  sulfur  ion  to  a 
sulfur  atom.  This  is  reduction  of  ferric  iron,  and-  the  sulfur 
has  been  oxidized. 

According  to  the  electronic  theory  an  atom  is  made  up 
of  a  positively  charged  nucleus  surrounded  by  negatively 
charged  corpuscles.  The  negative  corpuscles  are  called  elec- 
trons. In  the  ordinary  atom  the  amounts  of  positive  electricity 


OXIDATION  AND  REDUCTION  21 

and  negative  electricity  are  the  same.  If  an  atom  loses  an 
electron,  the  atom  will  then  carry  an  excess  of  positive  elec- 
tricity. The  loss  of  one  electron  will  result  in  the  atom's  hav- 
ing one  positive  charge.  For  example,  the  sodium  ion  is  a 
sodium  atom  minus  an  electron.  If  there  is  a  loss  of  two 
electrons  a  double  positive  charge  will  result,  as  in  the  cal- 
cium ion  (Ca++).  If  an  atom  gain  an  electron  it  will  become 
negatively  charged,  as  the  chlorine  ion.  If  sodium  chloride 
is  dissolved  in  water  an  electron  from  the  sodium  goes  to  the 
chlorine,  thus  giving  a  positive  sodium  ion  and  a  negative 
chlorine  ion.  The  ease  with  which  an  element  will  gain  or 
lose  an  electron  is  a  property  of  that  element.  Thus,  in  the 
oxidation  of  ferrous  chloride  to  ferric  chloride  by  chlorine, 
the  ferrous  ions  have  each  lost  an  electron  to  the  chlorine, 
which  gives  the  iron  an  added  unit  of  positive  electricity  and 
charges  the  chlorine  with  an  equal  amount  of  negative  elec- 
tricity. In  the  reduction  of  ferric  chloride  to  ferrous  chloride 
by  hydrogen  sulfide,  the  sulfur  ion  has  lost  two  electrons, 
and  each  ferric  ion  has  gained  one  from  the  sulfur.  A  study 
of  the  electromotive  series  of  metals  shows  that  for  the  most 
common  metals,  beginning  with  potassium,  the  tendency  to 
hold  electrons  increases  according  to  the  position  in  the  series : 
K,  Na,  Ba,  Sr,  Ca,  Mg,  Al,  Mn,  Zn,  Cd,  Fe,  Pb,  H,  Cu,  Hg, 
As,  Pt,  Ag.  Any  one  of  these  metals  will  take  electrons 
from  any  of  the  metals  preceding  it  in  the  series.  If  a  solu- 
tion of  copper  sulfate  is  treated  with  metallic  zinc,  the  copper 
ion  (Cu++)  will  take  two  electrons  from  the  zinc  atom,  caus- 
ing the  zinc  to  go  into  solution  as  the  zinc  ion  (Zn++),  while 
the  copper  will  be  reduced  by  these  electrons  and  be  precipi- 
tated as  metallic  copper: 

Zn  +  Cu+++  S04—  =  Zn+++  Cu  +  SO4~. 

The  nonmetallic  elements  as  a  rule  have  the  property  of 
taking  up  electrons  and  forming  negative  ions.  The  metallic 


22  NOTES  ON  QUALITATIVE  ANALYSIS 

elements  have  the  property  of  losing  electrons  and  forming 
positive  ions.  The  ionization  of  ionogens  in  solution  are 
examples  of  these  properties. 

It  is  not  necessary  that  oxidation  and  reduction  result  in 
the  formation  of  ions.  The  rusting  of  iron,  the  combination  of 
hydrogen  and  oxygen,  and  other  reactions  of  this  nature  are 
oxidation  reactions.  The  so-called  active  nonmetallic  elements 
—  those  that  have  a  very  strong  tendency  to  take  up  elec- 
trons, such  as  oxygen  and  chlorine  —  are  most  often  used 
as  oxidizing  agents.  The  following  equations  show  some  of 
the  substances  that  may  be  used  to  furnish  oxygen  and  the 
method  in  which  they  may  be  considered  to  decompose  to 
furnish  the  oxygen  : 

Na202  +  2  H2O  —  >-  2  NaOH  +  H2O2  ;  H2O2  —  *•  H2O  +  O. 
2  HN03  —  *  H20  +  2  NO  +  3  O. 
KC1O8 
Pb0 


KNO3  —  >•  KNO2  +  O. 

An  illustration  of  the  action  of  nitric  acid  as  an  oxidizing 
agent  is  its  effect  on  hydrochloric  acid.  A  mixture  of  these 
two  acids  is  called  aqua  regia: 

2  HNO8  —  >•  H2O  +  2  NO  +  3  O, 
6  HC1  +  3  O  —  H  3  H20  +  6  Cl. 

It  should  be  kept  in  mind  that  the  oxidation  of  a  substance 
cannot  take  place  without  the  reduction  of  other  substances. 
All  reactions  that  are  spoken  of  as  oxidation  reactions  are 
also  reduction  reactions.  The  substance  that  loses  the  elec- 
trons is  oxidized,  while  the  substance  that  takes  up  the 
electrons  is  reduced.  Most  oxidation  and  reduction  reactions 
are  conveniently  written  in  steps.  All  the  oxidation  reactions 
attempted  in  this  outline  will  be  written  in  the  simplest 


OXIDATION  AND  REDUCTION  23 

manner  in  which  they  may  be  considered  to  take  place. 
Whenever  possible,  the  oxidizing  agent  will  be  considered 
to  break  up  in  such  a  manner  as  to  furnish  free  oxygen, 
and  the  compound  to  be  oxidized  to  decompose  into  the 
oxide  of  the  element  concerned.  This  oxide  will  then  be 
acted  upon  by  the  free  oxygen  to  form  the  oxidized  product. 
To  illustrate  : 

When  manganese  hydroxide  is  oxidized  by  sodium  perox- 
ide the  reactions  may  be  assumed  to  be 

Na202  +  2  H20  —  K  2  NaOH  +  H2O2, 

H202  -  >-H20  +  0; 
Mn  (OH)2  -  >•  MnO  +  H2O  ; 
MnO  +  O  -  >•  MnO2. 

If  the  oxide  formed  is  an  anhydride  it  will  probably  combine 
with  water  to  form  an  acid  or  a  base.  For  example, 

S0  +  0  —  >-S0, 


If  it  is  a  hydrogen  compound  which  is  oxidized,  the  free 
oxygen  from  the  oxidizing  agent  will  usually  combine  with 
the  hydrogen,  forming  water,  thus  freeing  the  radical  com- 
bined with  the  hydrogen,  which  may  combine  with  other 
radicals  in  the  solution,  either  of  the  same  or  different  kinds. 
As  illustrations  may  be  cited  the  action  of  nitric  acid  on 
hydrochloric  acid  to  form  free  chlorine,  as  given  above,  two 
chlorine  atoms  uniting  to  form  a  molecule  ;  also  the  action 
of  nitric  acid  on  hydrogen  sulfide: 

2  HNO3  ^->-  H2O  +  2  NO  +  3  O, 
3  H2S  +  3  O  —  *  3  H20  +  3  S. 

If  this  reaction  is  considered  from  the  point  of  transfer  of 
charges,  it  is  seen  that  nitrogen  in  changing  from  valence  5 
in  HNO3  to  valence  2  in  NO  loses  three  positive  charges,  or 


24  NOTES  ON  QUALITATIVE  ANALYSIS 

2  N+++++ >-  2  N++  +  6  ®.    The  6  0  oxidizes  3  S~  to  free 

sulfur.    This  can  be  represented  as  follows : 

2  HNO8 +-  H2O  +2  NO  +  3  O~  +  6  ®  ; 

3  H2S *•  6  H+  4-  3  S~  ; 

6  H+  +  3  O~ >•  3  H2O  ; 

6  0  +  3  S—  — >•  3  S. 

The  oxygen,  being  charged  negatively,  combines  with  the  hy- 
drogen, which  is  positively  charged,  to  form  water.  Thus  the 
nitric  acid  has  been  reduced  by  giving  up  positive  charges, 
and  hydrogen  sulfide  oxidized  by  gaining  positive  charges. 


SOLUBILITIES 

A  general  knowledge  of  the  solubility  of  salts  and  other 
compounds  frequently  used  in  qualitative  analysis  is  of  great 
value  as  an  aid  to  predicting  when  precipitates  may  be 
formed  and  under  what  conditions  they  will  be  dissolved. 
It  will  also  often  be  a  great  aid  in  determining  possible  com- 
pounds in  a  mixture. 

The  following  are  given  as  a  few  simple  general  rules  of 
solubilities  of  compounds: 

All  sodium,  potassium,  and  ammonium  compounds  are 
soluble. 

All  normal  nitrates  are  soluble ;  all  normal  acetates  are 
soluble. 

All  chlorides  are  soluble  except  silver  chloride,  mercurous 
chloride,  and  lead  chloride.  The  latter  is  slightly  soluble  in 
cold  water  and  more  soluble  in  hot  water.  Some  basic  chlo- 
rides, notably  the  oxychlorides  of  antimony  and  bismuth,  are 
difficultly  soluble. 

All  sulfates  are  soluble  except  the  sulfates  of  lead,  ba- 
rium, strontium,  and  calcium,  the  latter  being  slightly 
soluble. 


GENERAL  INSTRUCTIONS  25 

With  few  exceptions  other  compounds  are  insoluble  in 
water.  These  include  oxides,  hydroxides,  carbonates,  sulfides, 
phosphates,  silicates,  tartrates,  oxalates,  etc. 

These  general  rules,  if  remembered,  will  often  be  very  use- 
ful in  the  course  of  an  analysis.  For  example,  if  barium  has 
been  found  in  an  unknown  mixture  which  is  soluble  in  water, 
the  sulfate  radical  cannot  be  present  and  need  not  be  tested 
for.  Similarly,  if  the  chloride  ion  is  found  in  a  soluble  mix- 
ture, silver  ions  cannot  be  present.  When  greater  detail  in 
regard  to  the  solubility  of  particular  compounds  is  desired, 
some  of  the  numerous  tables  of  solubilities  may  be  consulted. 

A  few  exceptions  to  these  general  rules  will  be  noted  when 
met  with  (see  Exp.  1,  Alkali  Group).  It  should  always  be 
borne  in  mind  that  even  soluble  substances  may  differ  very 
greatly  in  their  degree  of  solubility,  and  that  increase  in 
temperature  generally,  but  not  always,  increases  solubility. 
Hydrolysis  may  sometimes  cause  the  formation  of  basic  or 
acid  salts  which  show  a  different  solubility  from  the  corre- 
sponding normal  salts.  Examples  are  the  chlorides  of  anti- 
mony and  bismuth  and  the  sulfides  of  the  alkaline  earth  metals. 

GENERAL  INSTRUCTIONS 

Cleanliness  is  essential  to  success  in  chemical  analysis.  All 
apparatus  should  be  thoroughly  washed,  then  rinsed  with 
distilled  water  before  it  is  used  or  before  it  is  put  away. 
Observe  all  reagents  that  are  used,  and  if  not  clear  they 
should  be  filtered.  From  1  cc.  to  5  cc.  is  sufficient  unless  an 
excess  is  called  for.  In  neutralizing  solutions  or  making  them 
acid  or  alkaline,  great  care  should  be  taken  to  assure  thorough 
mixing  before  testing  with  litmus,  as  otherwise  only  a  part  of 
the  solution  maybe  neutralized.  Acidity  or  alkalinity  should 
never  be  taken  for  granted,  but  a  test  should  always  be  made. 
Prepare  a  wash  bottle  for  distilled  water,  and  keep  it  always 
filled  ready  for  use.  Distilled  water  must  be  used  exclusively 


26  NOTES  ON  QUALITATIVE  ANALYSIS 

except  in  washing  glassware.  Prepare  three  stirring  rods  by 
sealing  the  ends  of  a  glass  tube  about  the  length  of  a  lead 
pencil,  or  by  fire-polishing  the  end  of  a  glass  rod.  Also  prepare 
two  glass  tubes  about  six  inches  long  and  fire-polish  both  ends. 

Precipitation  is  effected  by  the  formation  of  an  insoluble 
compound  from  a  solution.  This  is  usually  accomplished  by 
adding  a  reagent  which  reacts  by  double  decomposition  to 
form  an  insoluble  product  with  the  ions  it  is  desired  to  re- 
move from  the  solution.  The  precipitated  solid  may  be  crys- 
talline or  amorphous.  It  may  rise  to  the  surface  of  the 
liquid  or  sink  to  the  bottom  of  the  vessel.  For  rapid  pre- 
cipitation the  solutions  should  be  stirred  and  heated  except 
where  heat  would  produce  undesired  reactions.  The  latter 
are  the  exception.  In  most  reactions  a  large  excess  of  the 
precipitant  should  be  avoided,  but  the  presence  of  a  slight 
excess  is  requisite  when  complete  precipitation  is  desired. 
To  prove  that  an  excess  of  the  precipitant  has  been  used,  let 
the  precipitate  settle,  or  filter  a  little,  and  add  a  few  drops  of 
the  precipitant  to  the  clear  solution.  If  a  precipitate  forms, 
not  enough  of  the  precipitant  has  been  added.  If  complete 
precipitation  is  not  desired,  but  formation  of  an  insoluble 
compound  only  to  be  observed,  addition  of  the  reagent  need 
not  be  continued  after  a  permanent  precipitate  forms. 

Filtration  consists  in  separating  a  liquid  from  a  solid  by  pour- 
ing the  mixture  on  a  filter  that  is  porous  enough  to  permit 
the  passage  of  the  liquid  but  will  retain  the  solid.  Sometimes 
the  solid  is  so  finely  divided  that  it  will  pass  through  the 
filter.  In  some  instances  this  may  be  remedied  by  heating  the 
mixture  for  some  time  and  filtering  through  two  thicknesses 
of  filter  paper.  Where  a  precipitate  is  not  crystalline,  rapid 
filtration  is  best  performed  by  allowing  the  precipitate  to 
settle  and  then  pouring  off  the  supernatant  liquid  as  com- 
pletely as  possible,  transferring  the  precipitate  to  the  filter 
afterwards.  The  process  of  pouring  off  the  supernatant  liquid 


GENEKAL  INSTRUCTIONS  27 

is  decantation.    Heating,  where  permissible,  will  often  help  to 
settle  a  precipitate  more  quickly  by  causing  coagulation. 

All  precipitates  collected  on  a  filter  are  washed  at  least 
twice  with  distilled  water  or  with  such  reagents  as  may  be 
called  for  in  the  instructions.  It  is  very  necessary  that  all 
precipitates  collected  on  a  filter  be  carefully  washed  with 
water  unless  there  are  specific  instructions  to  the  contrary. 
If  this  washing  is  not  thoroughly  carried  out  it  frequently 
happens  that  some  of  the  solution,  containing  metals  to  be 
tested  for  later,  will  be  retained  by  the  precipitate  in  suffi- 
cient quantity  to  interfere  seriously  with  the  identification 
of  metals  contained  in  the  precipitate.  Unless  otherwise 
stated,  the  washing  is  best  performed  with  hot  water.  In 
washing  a  filter  it  should  never  be  filled  more  than  two 
thirds  full  of  precipitate.  After  the  solution  has  drained  off, 
a  small  stream  of  water  from  the  wash  bottle  should  be 
directed  into  and  around  the  filter  just  below  the  top  in  such 
a  manner  as  to  force  the  precipitate  towards  the  point  of  the 
filter.  Great  care  should  be  used  never  to  add  enough  water 
to  fill  the  filter  to  the  top.  The  liquid  is  allowed  to  drain  out 
completely,  and  the  washing  repeated  as  often  as  necessary 
to  remove  all  soluble  material.  Time  may  often  be  saved  by 
allowing  the  precipitate  to  settle  and  pouring  off  the  superna- 
tant liquid  through  the  filter,  retaining  the  precipitate  in  the 
beaker.  Water  is  then  added  to  the  residue  in  the  beaker, 
and,  after  settling,  the  liquid  again  decanted  through  the 
filter.  This  may  be  repeated  three  or  four  times,  if  desired, 
before  the  residue  is  washed  into  the  filter.  All  precipitations 
with  hydrogen  sulfide  must  be  carried  out  in  a  closed  vessel.  A 
small  Erlenmeyer  flask,  with  a  one-hole  stopper  fitted  with  a 
six-inch  glass  tube,  serves  very  well.  For  precipitation,  remove 
the  stopper  and  allow  a  few  bubbles  of  hydrogen  sulfide  to 
pass  through  the  solution,  then  replace  the  stopper  and  shake 
the  vessel.  If  a  reaction  takes  place  the  gas  will  be  absorbed, 


NOTES  ON  QUALITATIVE  ANALYSIS 


causing  the  pressure  in  the  flask  to  decrease,  and  more  gas 
passes  into  the  solution  as  it  is  needed.  An  arrangement  of 
this  type  prevents  the  escape  of  the  gas  into  the  room. 

The  ability  to  make  a  satisfactory  analysis  depends  largely 
upon  being  able  to  recognize  a  reaction  by  the  color,  form  of 
precipitate,  etc.,  that  is  characteristic  of  the  different  sub- 
stances studied.  Description  alone  will  not  enable  one  to 
recognize  a  substance  or  a  reaction.  Preliminary  experiments 
are  given  which  are  illustrative  of  all  the  reactions  that  will 
take  place  in  the  analysis  of  an  unknown  substance  by  the 
outline.  Close  attention  must  be  paid  to  the  nature  of 
all  precipitates,  colors,  etc.,  and  these  observations  recorded 
briefly.  The  notes  should  be  recorded  in  a  tabular  form  and 
equations  written  for  all  reactions.  Below  is  an  example 
covering  a  few  experiments  on  the  alkaline  earth  group: 


CALCIUM 

BARIUM 

STRONTIUM 

MAGNESIUM 

Reagent 
NH4OH 

No  precipitate 

No  precipitate 

No  precipitate 

Mg(OH)2,  white, 
soluble  in  weak 
acids  and  ammo- 
nium chloride 

Reagent 
(NH4)2C08 

CaCO8,  white, 
insoluble  in 
NH4C1,  soluble 
in  HC2H802 

BaCO8,  white, 
insoluble  in 
NH4C1, 
soluble  in 
HC2H202 

SrCO3,  white, 
insoluble  in 
NH4Cl,soluble 
in  HC2H302 

MgC03Mg(OH)2, 
white,  soluble 
in  NH4C1  and 
HC2H302 

Flame 

Brick  red 

Green 

Crimson 

Etc. 

MgCl2 


Etc. 


2  NH4OH 

(NH4)2C08 
(NH4)2C03 
(NH4)2C08 


Mg(OH) 
CaCO 


2NH4Cl. 


2  NH4C1. 
SrC08  +  2NH4Cl. 
BaCO8  +  2  NH4C1. 


GROUPS  29 

In  the  study  of  qualitative  analysis  the  student  must 
constantly  remember  that  merely  following  the  directions  me- 
chanically brings  no  practical  knowledge  of  the  subject.  All 
experiments  and  tests  should  be  conscientiously  made  and 
all  observations  carefully  recorded.  Through  applying  the 
knowledge  of  chemistry  already  gained  strive  to  understand 
the  reason  for  each  step  and  to  interpret  and  write  each 
reaction.  Only  in  this  way  will  any  real  advance  be  made. 
In  the  laboratory  cleanliness,  patience,  and  care  will  be 
found  invaluable.  Strict  honesty  with  yourself  towards  your 
work  is  as  necessary  here  as  elsewhere.  The  student  who  is 
not  thoroughly  conscientious  will  gain  nothing.  Directions 
should  be  carefully  read  and  understood  before  beginning 
each  part  of  the  work  and  then  accurately  followed. 

For  the  purpose  of  qualitative  analysis  the  metals  are 
divided  into  groups,  each  group  including  metals  that  behave 
similarly  with  certain  reagents.  The  classification  is  based  on 
the  formation  of  insoluble  compounds  with  certain  reagents 
called  group  reagents.  The  following  is  the  classification 
used  in  this  outline : 

Silver  group.  Metals  whose  chlorides  are  insoluble  in  cold  water 
or  dilute  acids  :  lead,  silver,  and  mercurous  mercury. 

Arsenic  group.  Metals  whose  sulfides  are  insoluble  in  dilute 
acids  but  are  soluble  in  yellow  ammonium  sulfide :  arsenic,  anti- 
mony, and  tin. 

Copper  group.  Metals  whose  sulfides  are  insoluble  in  dilute 
mineral  acids  and  in  yellow  ammonium  sulfide :  copper,  bismuth, 
cadmium,  lead,  and  mercuric  mercury. 

Iron-zinc  group.  Metals  whose  sulfides  are  soluble  in  dilute 
mineral  acids  but  are  insoluble  in  ammoniacal  solution :  nickel, 
cobalt,  iron,  chromium,  aluminium,  manganese,  and  zinc. 

Alkaline  earth  group.  Metals  that  will  not  be  precipitated  by 
the  reagents  used  for  the  preceding  groups  but  will  be  pre- 
cipitated as  carbonates  in  the  presence  of  ammonium  chloride: 


30  NOTES  ON  QUALITATIVE  ANALYSIS 

barium,  calcium,  strontium.  Magnesium  will  also  be  considered 
with  this  group,  though  it  is  not  precipitated  by  the  carbonate 
ion  in  the  presence  of  much  ammonium  chloride. 

Alkali  group.  With  the  exception  of  magnesium,  metals  that 
will  not  be  precipitated  by  the  reagents  for  any  of  the  above 
groups  :  sodium,  potassium,  and  ammonium. 

THE  ALKALI  GROUP 

The  members  of  this  group  to  be  studied  are  sodium, 
potassium,  and  ammonium.  Almost  without  exceptions  the 
compounds  of  these  ions  are  soluble  in  water.  Ammonium 
is  classed  with  this  group  because  the  solubility  of  ammo- 
nium compounds  is  similar  to  that  of  sodium  and  potassium 
compounds.  The  identification  of  sodium  and  potassium  will 
depend  largely  upon  the  color  their  heated  vapors  impart 
to  a  flame.  Ammonium  compounds  are  all  decomposed  by 
heat  and  will  all  evolve  ammonia  when  heated  in  strong 
alkaline  solution.  On  account  of  the  solubility  of  these  com- 
pounds the  members  of  this  group  will  remain  in  solution 
while  all  others  are  removed  in  the  form  of  insoluble  salts. 

Use  solutions  of  sodium  chloride,  potassium  chloride,  and 
ammonium  chloride  in  the  experiments  which  follow. 

Experiment  1 

To  a  portion  of  each  solution  add  sodium  cobaltinitrite. 

Experiment  2 

Mix  a  small  amount  of  ammonium  chloride  solution  with 
sodium  hydroxide  solution  in  a  small  beaker.  Make  alkaline. 
Cover  the  beaker  with  a  watch  glass  on  the  under  side  of  which 
is  placed  >a  strip  of  moist  red  litmus  paper.  Warm  the  mixture 
but  do  not  boil.  Notice  the  change  in  color  of  the  litmus  and 
the  odor  of  the  evolved  gas.  Care  must  be  taken  to  prevent  any 
of  the  sodium  hydroxide  from  coming  in  contact  with  the  litmus 
paper  either  by  spattering  or  by  the  paper's  touching  the  sides  of 
the  beaker. 


ALKALINE  EARTH  GROUP  31 

Experiment  3 

Place  a  little  solid  ammonium  chloride  in  a  porcelain  crucible 
and  heat  to  redness  over  a  free  flame. 

Experiment  4 

Heat  in  a  colorless  flame  sodium,  potassium,  and  ammonium 
chloride  separately  by  dipping  in  solutions  of  each  a  clean 
platinum  wire  and  holding  the  wire  in  the  flame.  (Clean  the 
wire  carefully  each  time  before  placing  it  in  another  solution. 
This  may  be  accomplished  by  dipping  the  wire  in  concentrated 
hydrochloric  acid  and  holding  it  in  the  flame  until  the  flame  is 
colorless.)  Observe  the  flame  produced  by  each  through  three 
thicknesses  of  blue  glasses  or  a  color  screen.  Use  a  mixture  of 
sodium  chloride  and  potassium  chloride  and  observe  if  one  can 
be  detected  in  the  presence  of  the  other. 

THE  ALKALINE  EARTH  GROUP 

This  group  includes  calcium,  barium,  strontium,  and  mag- 
nesium. These  metals  occur  in  the  third  column  of  the  peri- 
odic table  and  are  uniformly  divalent  in  their  compounds. 
In  general  their  compounds  so  resemble  each  other  in  color 
and  solubility  that  the  separation  for  qualitative  analysis  will 
depend  largely  upon  the  degree  of  insolubility  of  certain 
characteristic  compounds.  This  group  will  remain  in  solution 
with  the  alkali  group  until  the  other  groups  have  been  pre- 
cipitated. Barium,  strontium,  and  calcium  are  precipitated 
from  magnesium  and  the  alkali  group  as  carbonates.  The 
table  on  page  32  gives  the  solubilities  of  some  of  the  more 
insoluble  salts  of  the  metals  of  this  group  together  with  the 
corresponding  sodium  and  potassium  compounds. 

The  upper  number  in  each  square  is  the  number  of  grams 
of  anhydrous  salt  held  in  solution  by  100  cc.  of  water  at  18°. 
The  lower  number  is  the  molar  solubility. 

The  table  of  solubilities  shows  the  hydroxides,  with  the 
exception  of  magnesium  hydroxide,  to  be  moderately  soluble 


32 


NOTES  ON  QUALITATIVE  ANALYSIS 


in  water.  Magnesium  hydroxide  will  be  precipitated  by 
ammonium  hydroxide,  while  the  other  members  of  the  group 
will  remain  in  solution.  This  method  for  the  separation  of 
magnesium  from  the  other  members  of  the  group  is  not 
practical,  because  precipitation  of  magnesium  as  hydroxide  is 
not  complete,  and  may  be  prevented  altogether  by  the  pres- 
ence of  ammonium  salts,  as  ammonium  chloride.  The  reason 


K 

Na 

Ba 

Sr 

Ca 

Mg 

OH 

142.9 
18 

116.4 
21 

3.7 
0.22 

0.77 
0.063 

0.17 
0.02 

0.0015 
0.0002 

SO4 

11.11 
0.62 

16.83 
1.15 

0.00023 
0.00001 

0.011 
0.0006 

0.20 
0.015 

35.43 

2.8 

Cr04 

63.1 

2.7 

61.21 
3.3 

0.00035 
0.000015 

0.12 
0.006 

0.4 
0.03 

73.0 
4.3 

<V>4 

30.27 
1.6 

3.34 
0.24 

0.0086 
0.00038 

0.0046 
0.00026 

0.00056 
0.000043 

0.03 
0.0027 

C08 

108.0 
5.9 

19.39 
1.8 

0.0023 
0.00011 

0.0011 
0.00007 

0.0013 
0.00013 

0.1 
0.01 

P04 

MgNH4P04 

0.00086 

for  this  interference  with  precipitation  by  ammonium  chlo- 
ride can  be  explained  by  the  solubility-product  principle  and 
the  effect  on  ionization  by  the  addition  of  a  common  ion.  It 
will  be  remembered  that  in  the  introduction  it  was  shown 
how  the  addition  of  an  ammonium  salt  to  a  solution  of 
ammonium  hydroxide  caused  a  marked  decrease  in  the  concen- 
tration of  hydroxyl  ions,  due  to  the  excess  of  ammonium  ions 
added.  A  0.1  molar  solution  of  magnesium  chloride  is  70 
per  cent  ionized,  thus  making  the  concentration  of  the  mag- 
nesium ion  0.07  molar.  The  solubility  product  of  magnesium 
hydroxide  is  c^  x  (^  =  3.4  x  10  -  » 

Substituting  the  value  of  the  magnesium  ion  in  0.1  molar 
solution  in  this  equation  and  solving,  the  concentration  of 


ALKALINE  EARTH  GROUP  33 

hydroxyl  ion  is  found  to  be  0.0000235  molar.  The  concen- 
tration of  hydroxyl  ions  must  exceed  this  value  in  order  that 
magnesium  hydroxide  be  precipitated  from  a  0.1  molar  solu- 
tion of  magnesium  chloride.  The  concentration  of  hydroxyl 
ions  in  a  solution  that  is  0.1  molar  with  respect  to  both 
ammonium  hydroxide  and  ammonium  chloride  is  0.000021. 
Solutions  of  these  concentrations  would  produce  no  precipi- 
tate of  magnesium  hydroxide.  Calcium,  barium,  and  stron- 
tium are  precipitated  as  normal  carbonates  by  ammonium 
carbonate  in  the  presence  of  ammonium  chloride.  Normal 
magnesium  carbonate  will  not  be  precipitated  by  ammonium 
carbonate  even  in  the  absence  of  ammonium  chloride,  because 
of  hydrolysis.  However,  a  basic  carbonate  of  varying  com- 
position may  be  precipitated  in  the  absence  of  ammonium 
chloride.  The  composition  of  the  precipitate  obtained  varies 
considerably  with  conditions  but  might  be  most  simply  written 

MgCOs .  Mg(OH)2. 

As  the  hydrolysis  is  an  equilibrium  reaction  represented  by 
the  equation, 

MgCO3  (solid)  ^i±  MgCO3  (dissolved) 
MgCO3  (dissolved)  +  2  HOH  +=±  Mg(OH)2  +  H2CO3, 

it  is  seen  that  the  addition  of  ammonium  chloride  would 
decrease  the  hydroxyl  ions  and  thus  prevent  the  precipita- 
tion of  the  magnesium  hydroxide.  This  will  cause  the  reac- 
tion to  proceed  more  rapidly  to  the  right,  and  the  magnesium 
carbonate  would  be  changed  into  magnesium  hydroxide, 
which  would  dissolve. 

As  the  carbonates  of  barium,  strontium,  and  calcium  are 
much  less  soluble  than  magnesium  carbonate,  and  are  not 
readily  dissolved  by  ammonium  chloride,  the  separation  of 
magnesium  from  the  other  members  of  the  group  will  be  by 
the  precipitation  of  the  carbonates  of  barium,  strontium,  and 
calcium  from  an  alkaline  solution  that  contains  ammonium 


34  NOTES  ON  QUALITATIVE  ANALYSIS 

chloride.  The  carbonates  of  barium,  strontium,  and  calcium 
are  slightly  soluble  in  ammonium  salts,  especially  when 
heated.  On  this  account  a  large  excess  of  ammonium  chlo- 
ride should  be  avoided  and  the  solution  should  not  be  boiled. 
The  carbonates  are  dissolved  in  acetic  acid.  Strong  acids 
must  not  be  used  because  they  would  interfere  with  the 
separation  of  barium  later.  Barium  is  separated  from  stron- 
tium and  calcium  by  precipitating  it  from  the  acetic-acid 
solution  as  barium  chromate  by  means  of  potassium  dichro- 
mate.  Potassium  dichromate  (K2Cr2O?)  is  the  acid  salt  of 
chromic  acid,  and  is  related  to  potassium  chromate  (K2CrO4) 
much  as  potassium  acid  sulfate  (KHSO4)  is  related  to  potas- 
sium sulfate  (K2SO4).  If  potassium  acid  sulfate  is  heated, 
the  following  reaction  takes  place: 


2KHS0 


Two  molecules  of  the  acid  salt  lose  one  molecule  of  water  to 
form  K2S2O7,  known  as  potassium  pyrosulfate.  When  to 
K2CrO4  an  acid  is  added,  KHCrO4  is  formed,  as  in  the  case 
of  potassium  acid  sulfate,  but  this  salt  loses  water  spontane- 
ously, according  to  2  KHCrO4^=±  K2Cr2O7  +  H2O,  to  form 
potassium  dichromate.  As  this  reaction  is  reversible  every 
solution  of  a  dichromate  and  every  acid  solution  of  a  chromate 
contains  both  CrO4~~  and  Cr2O7  ions.  When  a  barium  salt 
is  added  to  such  solution  the  Ba  ions  will  tend  to  combine 
with  both  the  CrO4~~  ions  and  the  Cr2O7~~  ions.  As  BaCrO4 
is  much  less  soluble  than  BaCr2O?,  the  CrO4~~  ions  are  con- 
stantly removed  from  the  solution  and  the  reaction  proceeds 
toward  the  left  until  all  the  dichromate  has  decomposed  into 
the  chromate  or  all  the  barium  has  been  precipitated.  Thus 
the  precipitate  formed  is  barium  chromate  (BaCrO4)  even 
when  a  dichromate  (K2Cr2O7)  is  used  as  the  reagent.  Strong 
acids,  by  their  tendency  to  form  dichromate  ions  (Cr2O7~~), 
will  prevent  the  precipitation  of  barium  as  chromate. 


ALKALINE  EARTH  GKOUP  35 

All  precipitates  that  form  in  this  group  should  be  further 
tested  by  the  flame.  It  must  be  remembered  that  the  sep- 
aration of  the  members  of  this  group  from  each  other  de- 
pends on  the  difference  in  solubility  of  similar  compounds, 
which  frequently  is  not  relatively  large,  and  as  concentrations 
often  cannot  be  known  it  sometimes  happens  that  the  precip- 
itate sought  may  be  mixed  with  that  of  some  other  member 
of  the  group.  In  particular  this  may  be  the  case  with  barium 
and  strontium  when  the  former  is  precipitated  as  chromate. 
Diluting  the  solution  decreases  the  possibility  of  precipitat- 
ing strontium  as  chromate.  To  insure  complete  separation  of 
barium  from  strontium  and  calcium,  the  solution  from  which 
the  barium  has  been  precipitated  as  chromate  is  treated  with 
ammonium  carbonate.  This  will  precipitate  the  strontium 
and  calcium  as  carbonates,  but  will  not  precipitate  any  barium 
left  in  solution.  The  carbonates  of  strontium  and  calcium 
are  much  less  soluble  than  the  corresponding  chromates ;  but 
barium  chromate  is  less  soluble  than  barium  carbonate  (see 
table  of  solubilities,  p.  32),  hence  there  are  too  few  barium 
ions  left  in  the  solution,  after  an  excess  of  chromate  ions  has 
been  added,  to  precipitate  with  the  carbonate  ion.  Upon  the 
same  principle  depends  the  identification  of  strontium  and  its 
separation  from  calcium.  Strontium  sulfate  is  so  much  less 
soluble  "than  calcium  sulfate  that  the  concentration  of  the 
sulfate  ion  in  a  saturated  solution  of  calcium  sulfate  will  be 
sufficient  to  precipitate  strontium  ions  even  from  a  fairly 
dilute  solution  of  a  strontium  salt.  The  sulfate  ions  being 
already  in  solution  in  combination  with  calcium  ions  would 
precipitate  no  calcium.  Strontium  and  barium  are  separated 
from  calcium  by  adding  sulfate  ions  in  moderate  concentra- 
tions, which  removes  practically  all  strontium  (and  barium) 
ions  from  solutions  as  sulfate.  Some  calcium  is  also  generally 
precipitated  as  sulfate,  but  a  saturated  solution  of  calcium 
sulfate  contains  calcium  ions  enough  to  precipitate  calcium 


36  NOTES  ON  QUALITATIVE  ANALYSIS 

oxalate  on  addition  of  the  oxalate  ion.  Calcium  oxalate  is 
much  less  soluble  than  calcium  sulfate  (see  table  of  solubili- 
ties, p.  32).  As  oxalic  acid  is  a  weak  acid,  the  presence 
of  any  strong  acid  tends  to  dissolve  the  calcium  oxalate  due 
to  the  combination  of  the  hydrogen  ions  with  the  oxalate  ion. 

Qalcium  cannot  be  tested  for  in  the  presence  of  strontium, 
either  by  the  flame  test  or  by  precipitation  as  oxalate.  As 
seen  from  the  table  (p.  32),  the  difference  in  solubility  of 
calcium  and  strontium  oxalates  is  so  slight  that  it  would  be 
impractical  to  attempt  to  prevent  the  precipitation  of  stron- 
tium with  calcium  by  controlling  the  concentration  of  the 
oxalate  ion ;  hence  the  separation  of  the  two  is  necessary. 

Strong  acids  must  not  be  used  for  dissolving  the  carbon- 
ates, because  barium  is  to  be  precipitated  as  chromate,  and 
the  salt  cannot  be  precipitated  in  the  presence  of  a  strong  acid : 

2  BaCr04  +  4  HC1  =  H2Cr2O7  +  2  Bad,  +  H2O. 

Use  solutions  of  the  nitrates  or  chlorides  of  barium,  stron- 
tium, calcium,  and  magnesium  in  the  experiments  which  follow. 

Experiment  5 

Make  each  solution  alkaline  with  ammonium  hydroxide,  then 
add  ammonium  carbonate.  Add  to  each  resulting  mixture  enough 
acetic  acid  to  give  the  solution  an  acid  reaction.  Could  these  pre- 
cipitates be  formed  in  acid  solution  ? 

Experiment  6 

Add  to  each  solution  an  equal  volume  of  ammonium  chloride 
solution,  make  alkaline  with  ammonium  hydroxide,  and  add  am- 
monium carbonate.  What  difference  do  you  observe  between 
these  results  and  those  in  Experiment  5  ?  Filter  each  mixture  and 
wash  the  precipitates  in  a  little  warm  water  by  pouring  the  water 
over  the  precipitate  collected  on  the  filter  and  letting  it  pass 
through  the  filter.  Dissolve  each  precipitate  in  the  least  amount 
of  acetic  acid  possible  by  pouring  the  acid  over  the  precipitate  on 
the  filter.  Add  a  few  drops  of  five  per  cent  potassium  clicliromate 


ALKALINE  EAKTH  GKOUP  37 

solution  to  each  nitrate.    Test  the  solubility  of  any  precipitate 
that  forms  in  hydrochloric  acid. 

Experiment  7 

Heat  solutions  of  each  almost  to  boiling,  then  add  five  per  cent 
ammonium  sulfate  solution,  a  few  drops  at  a  time,  until  an  equal 
volume  has  been  added.  Let  the  mixtures  stand  for  ten  minutes, 
filter,  make  the  filtrates  alkaline  by  addition  of  ammonium  hydrox- 
ide, and  then  add  ammonium  oxalate.  (See  table  of  solubilities 
for  explanation  of  results.)  Test  the  solubility  of  the  precipitate 
in  acetic  acid  and  in  hydrochloric  acid. 

Experiment  8 

To  each  solution  add  an  equal  volume  of  a  saturated  solution 
of  calcium  sulfate.  Heat  to  boiling  and  let  stand  for  ten  minutes. 

Experiment  9 

To  a  solution  of  magnesium  chloride  or  nitrate  add  five  times 
its  volume  of  water,  enough  ammonium  chloride  to  prevent  the 
precipitation  of  magnesium  hydroxide,  make  alkaline  with  am- 
monium hydroxide,  and  add  disodium  phosphate  solution.  The 
precipitate  formed  should  be  crystalline.  If  no  precipitate  forms 
at  first,  shake  vigorously  and  let  stand  for  a  few  minutes. 

Experiment  10 

Observe  the  color  imparted  to  a  colorless  flame  by  the  heated 
vapors  of  the  chlorides  of  barium,  strontium,  and  calcium.  (See 
Experiment  4,  p.  31.)  Observe  the  colors  with  and  without  the 
blue  glasses  or  color  screen. 

NOTES  ON  ANALYSIS 

1.  The  flame  test  is  best  made  in  hydrochloric  acid  solution, 
because  the  chlorides  are  more  volatile  than  many  other  salts. 
The  solutions  should  be  concentrated. 

2.  In  dilute  solutions  formation  of  a  precipitate  may  be  very 
slow  on  account  of  the  formation  of  supersaturated  solutions.    In 
the  precipitations  of  this  group  considerable  time  should  be  al- 
lowed before  discarding  the  solutions.    If  no  precipitate  forms  at 
first,  shake  vigorously.  Warming  sometimes  hastens  precipitations. 


38 


NOTES  ON   QUALITATIVE  ANALYSIS 


3.  When  the  alkaline  earth  metals  are  to  be  precipitated 
from  a  filtrate  from  which  other  groups  have  been  removed,  it 
will  generally  be  best  to  concentrate  the  solution  before  adding 
ammonium  carbonate. 


SCHEME  FOR  ANALYSIS  OF  THE  ALKALI  AND  THE  ALKALINE 
EARTH  GROUPS 


Solution  contains  Ca++,  Ba++,  Sr++,  Mg++,  NH4+,  K+,  Na+ 

a.  Test  a  small  portion  of  the  solution  for  ammonium  according 
to  Experiment  2. 

b.  Try  the  flame  test  for  sodium. 

c.  To  a  larger  portion  of  the  solution  add  NH4C1  until  addition  of 
NH4OH  gives  no  precipitate.   Make  alkaline  with  NH4OH  and  add 
(NH4)2CO8.    Heat  to  60°-70°  for  a  few  minutes,  adding  a  few  drops 
of  (NH4)2CO8  from  time  to  time.     Filter,  and  test  filtrate  with 
(NH4)2CO8  to  make  sure  of  complete  precipitation.   Wash  the  pre- 
cipitate with  hot  water. 


Precipitate  :  BaC03,  CaC03,  SrC03 
Dissolve  in  the  smallest  possible 
amount  of  hot  dilute  acetic  acid 
by  pouring  the  acid  repeatedly 
over  the  precipitate  collected  on 
the  filter.  Test  a  small  portion 
of  the  solution  for  Ba  by  adding 
a  few  drops  of  K2O2Or  Warm 
the  solution.  If  Ba  is  present, 
a  precipitate  will  be  formed.  If 
Ba  is  absent,  dilute  all  of  the 
acid  solution  to  about  10  cc.,  di- 
vide into  two  parts,  and  proceed 
with  the  analysis  for  Sr  and  Ca. 
If  Ba  is  present,  dilute  all  of 
the  acid  solution  to  about  five 
times  its  volume.  Warm  and  add 
K2Cr2O7  until  the  solution  is 
distinctly  yellow.  Filter,  and 
wash  the  precipitate  with  warm 
water.  Analyze  precipitate  and 
filtrate  as  directed  on  page  39. 


Filtrate 

Add  a  few  drops  of  (NH4)2SO4 
and  (NH4)2C2O4;  warm  and  fil- 
ter if  necessary.  Divide  the  solu- 
tion into  two  parts.  (1)  Acidify 
with  HC1.  Evaporate  to  dryness 
and  ignite  (heat  strongly)  until 
ammonium  salts  are  entirely  re- 
moved. Dissolve  the  residue  in 
3~4  cc.  of  water  to  which  a  few 
drops  of  HC1  have  been  added.  Fil- 
ter if  not  clear  and  to  the  solution 
add  Na8Co(NO2)6.  Confirm  a  pre- 
cipitate by  the  flame  test.  (2)  To 
the  second  part,  made  alkaline  with 
NH4OH,  add  Na2HP04.  A  white 
crystalline  precipitate  indicates 
Mg.  Confirm  by  dissolving  in  the 
least  amount  of  HC1  possible,  heat 
to  boiling,  and  neutralize  with 
NH4OH.  The  precipitate  should 
reappear  in  a  crystalline  form. 


ALKALINE  EARTH  GKOUP 


39 


SCHEME  FOR  ANALYSIS  OF  THE  ALKALI  AND  THE  ALKALINE 
EARTH  GROUPS  (Continued  from  page  38) 


Precipitate :  BaCrO* 
Dissolve  in  a  few 
drops  of  concen- 
trated HC1  and  con- 
firm by  the  flame 
test.  Green  flame 
proves  Ba. 


Filtrate 

Make  alkaline  with  NH4OH,  add  (NH4)2CO3, 
warm  to  60°-70°,  for  a  few  minutes,  filter,  and 
reject  the  filtrate.  Wash  the  precipitate,  dis- 
solve in  acetic  acid,  dilute  to  10  cc.,  and  divide 
into  two  parts.  (1)  Test  one  part  for  Sr  by 
adding  an  equal  volume  of  CaSO4  solution. 
Heat  and  let  stand  for  at  least  ten  minutes.  If 
a  precipitate  forms,  confirm  Sr  by  the  flame  test. 
(2)  Test  the  second  portion  for  Ca.  If  Sr  is  pres- 
ent, make  the  solution  alkaline  with  NH4OH 
and  add  an  equal  volume  of  (NH4)2SO4.  Heat 
for  a  few  minutes,  let  cool,  and  filter.  Make 
the  filtrate  alkaline  with  NH4OH  and  add 
(NH4)2C2O4.  A  precipitate  insoluble  in  acetic 
acid  is  probably  CaC2O4.  Confirm  by  the 
flame  test.  If  Sr  is  not  present  the  treatment 
with  (NH4)2SO4  is  omitted. 


4.  No  precipitates  will  be  produced  in  solutions  of  calcium, 
strontium,  and  barium  salts  on  the  addition  of  ammonium  hydrox- 
ide if  this  reagent  is  free  from  carbonates.   If  carbonate  is  present 
in  the  ammonium  hydroxide,  however,  a  turbidity  may  result  due 
to  the  formation  of  the  carbonates  of  these  metals.   This  turbidity 
may  also  appear  when  solutions  of   these  metals,  after  having 
been  made  alkaline  by  the  addition  of  ammonium  hydroxide,  are 
allowed  to  stand,  on  account  of  the  absorption  of  carbon  dioxide 
from  the  air. 

5.  The  identification  of  the  members  of  this  group  is  made 
more  certain  if  the  spectroscope  is  used  in  connection  with  the 
flame  reactions.    By  this  means  the  presence  of  the  metals  in 
quantities  too  small  to  produce  conclusive  precipitates  may  be 
proved.   These  tests  have  not  been  included  in  the  outline  because 
it  is  often  not  feasible  to  use  the  spectroscope  with  large  classes. 


40  NOTES  ON  QUALITATIVE  ANALYSIS 

THE  SULFIDE  GROUPS 

Those  metals  which  in  this  outline  are  separated  from  the 
metals  of  other  groups  by  their  precipitation  as  sulfides  in- 
clude mercury  (mercuric),  lead,  bismuth,  copper,  cadmium, 
arsenic,  antimony,  tin,  iron,  cobalt,  nickel,  zinc,  and  manga- 
nese. Aluminium  and  chromium  are  precipitated  in  these 
groups  but  on  account  of  hydrolysis  are  not  precipitated  as 
sulfides  but  as  hydroxides.  If  the  position  of  these  elements 
in  the  periodic  table  is  noted,  it  is  observed  that  they  are 
scattered  through  groups  I  to  VIII  inclusive  and  will  be 
expected  to  vary  greatly  in  their  chemical  conduct.  Some 
are  moderately  strong  metals,  while  others  show  pronounced 
nonmetallic  properties.  Most  of  them  show  more  than  one 
valence,  with  characteristic  properties  for  each.  Many  enter 
easily  into  complex  ions,  while  others  show  little  such  tend- 
ency and  behave  normally  in  this  respect. 

This  great  variation  in  chemical  behavior  makes  it  possible 
and  necessary  to  subdivide  further  these  metals,  after  precip- 
itation, into  smaller  groups,  in  which  identification  is  more 
easily  carried  out. 

The  reagent  used  for  the  precipitation  of  these  metals  is 
hydrogen  sulfide.  The  separation  into  subdivisions  depends 
primarily  upon  the  concentration  of  the  sulfide  ion  necessary 
for  the  precipitation  of  the  different  metal  ions.  This,  in 
turn,  depends  upon  the  concentration  of  hydrogen  ion  in  the 
solution,  or,  in  other  words,  the  acidity  or  alkalinity  of  the 
solution,  which  controls  to  a  very  large  extent  the  concen- 
tration of  sulfide  ion  which  can  be  present  in  a  saturated 
solution  of  hydrogen  sulfide.  To  understand  the  principle 
of  this  separation  a  careful  study  of  the  effect  which  varia- 
tion in  the  acidity  of  the  solution  (concentration  of  H  ion) 
will  have  upon  the  sulfide  ion  is  therefore  of  the  greatest 
importance  to  the  student.  . 


THE  SULFIDE  GROUPS  41 

Hydrogen  sulfide   is  a  dibasic  acid  and  shows  a  primary 
and  secondary  ionization.    The  primary  ionization, 


shows  an  ionization  constant 
X  CHS 


and  the  secondary  ionization, 

shows  the  constant 

xC«- 


In  a  saturated  solution  of  hydrogen  sulfide  the  concentration 
would  be  very  nearly  constant,  and  at  25°  and  atmospheric 
pressure  CH2S  =  0.1  molar.  Substituting  this  value  in  equa- 

1  (1)'  (3)  CH+  X  CHS-  =  0.91  x  10-8. 

As  the  molar  concentrations  of  the  H+  ions  and  HS~  ions 
are  equal,  the  value  for  each  would  be 


V0.91  x  10-8  =  0.95  x  10-4. 
Multiplying  equations  (1)  X  (2), 

2+  X  C-  _ 


Substituting  in  this  the  value  for  CH!!S  in  a  saturated  solution, 

(5)  C2H+xCs-  =  l.lxlO-23. 
Substituting  in  this  the  value  for  CH+  found  above, 

1  1  v  1  n  -23 
<6>C«—  (.96x10-0'  -1-2Xl°"" 


42 


NOTES  ON  QUALITATIVE  ANALYSIS 


This  last  value  is  the  concentration  of  sulfide  ion  (S — )  in  a 
solution  saturated  with  H2S  at  25°  and  atmospheric  pres- 
sure.* If  any  acid  is  added  to  a  saturated  solution  of  hydro- 
gen sulfide,  the  large  increase  in  the  hydrogen  ions  in  solution 
would  cause  a  decrease  in  the  concentration  of  the  sulfur 
ions.  The  value  for  the  concentration  of  the  sulfur  ions  may 
be  found  by  substituting  the  molar  concentration  of  the 
hydrogen  ion  in  equation  (5). 

The  following  table  shows  the  values  in  a  saturated  hydro- 
gen-sulfide  solution  for  the  concentrations  of  the  sulfur  ion 
in  the  presence  of  varying  amounts  of  hydrochloric  and 
acetic  acids: 


MOLECULAR 
CONCENTRA- 
TION OF  ACID 

PER  CENT  HC1 
DISSOCIATED 

C§  —  IN  PRES- 
ENCE OF  HC1 

PER  CENT 
ACETIC  ACID 
DISSOCIATED 

CH  —  IN  PRES- 
ENCE OF 
ACETIC  ACID 

0. 

1.2  X  10~15 

1.2  X  10-  I* 

0.001 

98.7 

1.1  x  lO-" 

11.62 

3.2  x  10-  16 

0.01 

96.6 

1.18  x  10-19 

4.05 

6.7  x  10~17 

0.1 

91.56 

1.6  x  10-21 

1.303 

6.5  x  10-  18 

0.2 

89.3 

3.4  x  10-22 

0.917 

0.6 

85.38  ' 

6.1  x  10-23 

0.5694 

1.3  x  10-  18 

From  the  table  it  is  seen  that  the  concentration  of  the  sul- 
fur ion  is  approximately  one  millionth  (10~6)  as  large  in  the 
presence  of  0.1  molar  hydrochloric  acid  as  in  neutral  solu- 
tion. Since  acetic  acid  is  very  slightly  ionized,  it  causes  very 
little  change  in  the  concentration  of  the  sulfur  ion.  The  fol- 
lowing table,  taken  from  Treadwell  and  Hall's  "  Qualitative 
Analysis,"  gives  the  solubility  and  solubility  product  for  the 
sulfides  of  a  number  of  metals: 


*See  Stieglitz,  Qualitative  Analysis,  Vol  I,  pp.  199-201. 


THE   SULFIDE  GROUPS 


43 


SUBSTANCE 

GRAMS  PER 
LITER 

MOLES  PER 
LITER 

SOLUBILITY  PRODUCT 

CuS     .     .     . 

8.8  x  10-21 

8.2  x  10-23 

CCu++xCs—  =  8.5xlO-45 

CdS     .     .     . 

8.6  x  10-  13 

6.0  X  lO-w 

Ccd++  X  Cs—  =  3.6  x  10-29 

FeS      .     .     . 

3.4  x  10-8 

3.9  x  10-10 

(W  +  xCs—  =  1.5xlO-i9 

HgS    .     .    . 

CHg++  x  Cs—  =4.0  x  10-  53 

MnS    .     .     . 

3.3  x  10-6 

3.8  x  10-8 

CMn++xCs—  =1.4x10-15 

NiS      ... 

7.0  x  10-n 

1.2  x  10-12 

CNi++  x  Cs—  =  1.4x  10-2* 

PbS     .     .     . 

4.9  x  10-12 

2.0  x  10-14 

CPb++  x  Cs—  =4.2  x  10-  28 

ZnS     .     .     . 

3.3  x  10-io 

3.5  x  10-12 

CZn++  x  Cs—  =  1.2  x  10-28 

H2S     .     .     . 

CH+  x  Cs—  =1.1  x  TO-23 

By  substituting  in  the  solubility-product  equation  for  any 
metallic  sulfide  the  value  of  Cs__  found  above,  the  concen- 
tration of  the  metallic  ion,  which  must  be  present  in  order  to 
cause  precipitation,  may  be  calculated. 
To  illustrate : 


CMn++xCs-=1.4xlO-15; 


then 


1.4  xlO-1 
1.2xlO-15 


=  1.16  moles. 


This  means  that  in  a  neutral  solution  saturated  with  hydro- 
gen sulfide  the  molar  concentration  of  the  manganese  ion 
must  be  1.16  moles,  which  is  equivalent  to  1.16  x  54.93,  or 
63.7  grams  per  liter,  before  a  precipitate  of  manganese  sul- 
fide will  be  produced.  A  similar  calculation  will  show  that 
in  a  neutral  solution  only  1  x  10  ~8  moles,  or  6.5  x  10 "7  grams, 
of  zinc  ion  must  be  present  in  order  that  a  precipitate  of  zinc 
sulfide  may  be  produced.  For  cadmium  ion  3  x  10~14  moles, 
or  3.4  x  10~12  grams  per  liter,  is  all  that  would  be  required 
to  produce  a  precipitate  of  cadmium  sulfide. 


44  NOTES  ON  QUALITATIVE  ANALYSIS 

Making  use  of  the  values  for  Cg--  in  the  table  on  page  42, 
it  is  easy  to  determine  the  concentration  of  any  metal  ion 
necessary  to  produce  a  precipitate  in  solutions  of  various 
strengths  of  acid.  To  illustrate:  The  solubility  product  for 
ZnS  is  CZn++  x  C8—  =1.2  x  10 ~23.  Substituting  the  value  of 
Cs--  for  0.2  molar  HC1  in  this  equation, 

1  9  v  1  0  ~23 
Czn  =  34*10-23  =  0-34  x  10 -1,  or  0.034  moles ; 

or  0.034  x  65  =  2.2  grams  per  liter. 

The  precipitation  of  zinc  sulfide,  even  from  a  solution  which 
was  neutral  in  the  beginning,  is  never  complete,  however, 
for  the  hydrochloric  acid  which  accumulates  from  the  reac- 
tion ZnCl2  +  H2S >•  ZnS  +  2  HC1  furnishes  many  hydrogen 

ions,  which  soon  greatly  decrease  the  value  of  Cs-  -,  as  can  be 
seen  from  the  table,  and  bring  the  precipitation  to  a  stand- 
still. By  a  similar  calculation  for  cadmium  sulfide  in  0.2 
molar  hydrochloric  acid  solution,  the  value  for  Ccd++  neces- 
sary to  produce  a  precipitate  can  be  shown  to  be  1  xlO~7x 
112,  or  0.00001  grams  per  liter.  So  if  the  concentration  of  the 
acid  did  not  get  above  0.2  molar,  cadmium  would  be  practi- 
cally all  precipitated  from  a  solution  of  any  of  its  salts  in  a 
solution  saturated  with  H2S.  These  figures  show  that  the 
ratio  between  the  quantities  of  cadmium  ion  and  zinc  ion 
necessary  to  produce  a  precipitate  in  0.2  molar  solution  of 
HC1  is  1 :  22000.  Lead  sulfide  and  tin  sulfide  show  about 
the  same  solubility  as  cadmium  sulfide,  these  three  being  the 
most  soluble  sulfides  of  the  copper  and  arsenic  groups.  As 
zinc  sulfide  is  the  least  soluble  sulfide  of  the  iron-zinc  group, 
it  is  easily  seen  that  in  a  solution  having  acid  concentration 
sufficient  to  prevent  the  precipitation  of  zinc  sulfide,  —  and 
hence  sulfides  of  all  other  metals  of  the  iron  group,  —  cadmium, 
lead,  and  tin  sulfides  would  be  practically  completely  pre- 
cipitated along  with  the  other  sulfides  of  the  copper  and 


THE  SULFIDE  GROUPS  45 

arsenic  groups.  Regulating  the  acid  concentration,  therefore, 
furnishes  a  convenient  method  for  the  separation  of  the  copper 
and  arsenic  groups  from  the  iron-zinc  groups,  and  is  the  basis 
for  this  division.  In  practice  it  is  found  that  an  acid  con- 
centration of  0.25  molar  gives  a  very  complete  separation  of 
these  groups.  The  approximate  concentration  of  hydrochloric 
acid  in  any  solution  may  be  determined  by  means  of  methyl 
violet,  as  will  be  illustrated  later. 

The  methods  for  separation  of  the  groups  from  each  other 
have  so  far  been  based  chiefly  upon  physical  differences  be- 
tween the  groups.  Advantage  is  often  taken  of  the  relative 
difference  in  solubility  of  compounds,  as,  for  example,  the 
difference  in  solubility  of  the  carbonates  of  the  alkali  and 
alkaline  earth  groups  and  the  difference  in  solubility  of  the 
sulfides  of  the  copper  and  the  iron-zinc  groups. 

In  the  separation  of  the  arsenic  group  from  the  copper 
group  use  is  made  of  chemical  differences  between  the  ele- 
ments in  the  two  groups.  So  far  the  elements  studied  have 
been  decidedly  base-forming  in  their  properties.  If  the  posi- 
tion of  arsenic,  antimony,  and  tin  in  the  periodic  table  be 
noted,  it  is  seen  that  they  are  probably  more  acid-forming 
than  base-forming  in  their  properties.  A  study  of  these  ele- 
ments proves  this  to  be  the  case.  As  would  be  expected,  the 
acid-forming  properties  are  especially  prominent  in  the  higher 
valences  of  these  elements.  The  lower  oxides  all  show  some 
basic-forming  tendency  (particularly  is  this  true  of  tin),  and 
even  in  the  higher  oxides  are  seen  some  base-forming  prop- 
erties, which,  however,  are  very  slight.  All  of  these  elements 
form  chlorides  which,  while  often  much  hydrolyzed,  neverthe- 
less give  enough  positive  ions  in  hydrochloric  acid  solution 
to  be  precipitated  by  hydrogen  sulfide.  Arsenic  is  the  most 
strongly  acid-forming  of  this  group,  and  tin  the  least.  It  is  a 
well-known  fact  that  oxygen  may  be  replaced  by  sulfur  without 
any  profound  change  in  the  chemical  nature  of  the  compound. 


46  NOTES  ON  QUALITATIVE  ANALYSIS 

It  would  be  expected,  then,  that  the  sulfides  of  this  group 
would  show  acid  properties.  Advantage  is  taken  of  this  prop- 
erty to  separate  these  elements  from  the  copper  group.  Just  as 
carbon  dioxide  will  combine  with  calcium  oxide  to  form  cal- 
cium carbonate,  so  carbon  disulfide  combines  with  potassium 
sulfide:  C02  +  CaO-+CaC03 

CS2  +  K2S — ^K2CS3. 

In  the  same  way  the  sulfides  of  this  group,  particularly 
the  higher  ones  (As2S5,  Sb2S6,  SnS2),  and  indeed  all  of  the 
lower  sulfides  except  SnS,  will  'combine  with  sulfides  of  the 
alkali  metals  to  form  soluble  salts  of  the  thio  acids: 

As2S8  +  3  (NH4)2S  — ^  2  (NH4)8AsS8. 
This  salt,  ammonium  thioarsenite,  ionizes  as  follows: 
(NH4)3AsS8  ^=±  3  NH4+  +  AsS,— . 

Stannous  sulfide  is  too  base-forming  to  produce  thio  salts 
in  this  way.  It  is  therefore  necessary  to  sulfurize  it  before 
a  complete  separation  of  the  arsenic  group  from  the  copper 
group  can  be  accomplished.  This  is  brought  about  by  the 
use  of  yellow  ammonium  sulfide,  which  contains  polysulfides, 
as  (NH4)2S2  etc.  The  action  may  be  represented  as  follows : 

SnS  +  (NH,)2S2  — -f  SnS2  +  (NH4)2S 
SnS,  +  (NH4)aS— »- (NH4)2SnS8. 

The  free  thio  acids  from  which  these  salts  are  derived 
are  very  unstable,  breaking  up  when  freed,  just  as  carbonic 
acid  does.  If  a  strong  acid  be  added  to  a  solution  containing 
a  thio  salt,  the  sulfide  is  precipitated  as  represented  by  the 
following  equations: 

(NH4)3  AsS8  +  3  HC1  — *  H3AsS3  +  3  NH4C1 
2HAsS— 


SEPARATION  OF  THE  SULFIDE  GROUPS         47 

SEPARATION  OF  ARSENIC,  COPPER,  AND  IRON  GROUPS 
FROM  EACH  OTHER.    REACTIONS  OF  SULFIDE  ION* 

Experiment  11 

1.  Place  about  5  cc.  of  a  solution  of  mercuric  chloride  in  a  pre- 
cipitation flask,  add  a  few  drops  of  dilute  hydrochloric  acid,  heat 
to  boiling,  and  saturate  with  hydrogen  sulfide.    Test  the  solubility 
of  the  precipitate  in  yellow  ammonium  sulfide.    To  do  this,  filter 
a  little  of  the  solution  and  test  the  precipitate  by  transferring  a 
little  of  it  to  a  test  tube  and  adding  1  cc.  of  yellow  ammonium 
sulfide  solution.   Warm  if  necessary,  filter,  and  acidify  the  filtrate 
with  dilute  HC1. 

2.  Repeat  1,  using  instead  of  mercuric  chloride  each  of  the 
following  solutions :  lead  acetate,  bismuth  chloride,  copper  sul- 
fate, cadmium  sulfate,  sodium  arsenite,  antimony  chloride,  stan- 
nous  chloride. 

Experiment  12 

1.  Place  about  5  cc.  of  each  of  the  following  solutions  succes- 
sively in  a  precipitation  flask,  add  a  few  drops  of  dilute  hydro- 
chloric acid,  warm,  and   saturate  with  hydrogen  sultide  as  in 
Experiment  11 :  ferric  chloride,  potassium  alum,  chrome  alum, 
zinc  sulfate,  cobalt  chloride,  nickel  chloride,  manganese  sulfate. 

Does  a  precipitate  form  in  any  case  ? 

2.  Using  solutions  of  the  same  salts  as  in  1,  treat  each,  without 
acidifying,  with  a  few  drops  of  ammonium  sulfide  solution.   Note 
if  a  precipitate  is  formed  and  its  color.    Save  the  solutions  con- 
taining the  precipitates. 

3.  Without  filtering  or  heating,  add  to  each  precipitate  obtained 
in  2  an  equal  volume  of  dilute  hydrochloric  acid.    If  the  sulfides 
do  not  dissolve  readily,  leave  them  in  contact  with  the  acid  three 
or  four  minutes. 

4.  Tabulate  the  results  obtained  in  1.    Would  you  expect  all  sul- 
fides formed  in  2  to  be  soluble  in  dilute  acid  ?  Explain  the  results 
obtained  in  3.  What  use  may  be  made  of  these  results  in  analysis  ? 

*  If  time  does  not  permit  a  study  of  all  the  ions  in  these  groups,  the  in- 
structor may  shorten  the  work  by  using  salts  of  a  few  of  the  typical  ions  in 
each  group,  omitting  such  others  as  may  seem  desirable. 


48  NOTES  ON  QUALITATIVE  ANALYSIS 

The  arsenic  and  copper  groups  are  best  separated  from  the 
iron-zinc  group  by  precipitating  them  in  a  hydrochloric-acid 
solution  0.25  to  0.30  molar.  As  mentioned  above,  this 
strength  may  be  determined  by  means  of  methyl  violet. 

Methyl  violet  in  the  presence  of  concentrated  hydrochloric 
acid  is  yellow,  but  in  the  presence  of  very  dilute  acid  it  is 
blue.  As  the  acid  concentration  becomes  less,  the  color  of 
the  methyl  violet  changes  successively  as  follows:  yellow, 
greenish-yellow,  green,  greenish-blue,  blue.  The  concentra- 
tion of  acid  which  gives  a  green  color  with  methyl  violet, 
without  a  tinge  of  yellow  or  blue,  represents  a  concentration 
of  0.25  molar  hydrochloric  acid,  which  is  the  concentration 
of  acid  to  make  a  complete  precipitation  of  all  the  metals 
in  the  above  list,  from  arsenic  to  stannous  tin  inclusive. 

Experiment  13 

Make  a  number  of  marks  on  white  porcelain  (or  a  piece  of 
white  paper)  with  an  indelible  pencil  (methyl  violet).  Place  1  cc. 
of  concentrated  hydrochloric  acid  in  a  beaker.  By  means  of  a 
stirring  rod  moisten  one  of  the  pencil  marks  with  one  drop  of  the 
acid.  Note  the  color.  Now  dilute  the  1  cc.  of  concentrated  hydro- 
chloric acid,  first  by  adding  5  cc.  of  water,  and  then  add  succes- 
sively the  following  amounts  :  6  cc.,  12  cc.,  24  cc.,  48  cc.  After 
each  dilution  moisten  one  of  the  indelible  pencil  marks  with  one 
drop  of  the  acid.  If  the  concentrated  hydrochloric  acid  were 
12  molar,  what  would  be  the  molar  Concentration  after  each 
dilution  ? 

Compare  the  colors  of  the  methyl  violet  produced  by  each  molar 
concentration  of  acid.  This  test  should  be  used  whenever  the 
arsenic  and  copper  groups  are  to  be  separated  from  the  iron-zinc 
group. 


AMPHOTERIC  HYDROXIDES  49 

THE  IRON-ZINC  GROUP 
AMPHOTEKIC  HYDROXIDES 

If  a  solution  of  sodium  hydroxide  is  added  to  a  solution 
of  a  zinc  salt,  a  precipitate  of  zinc  hydroxide  will  be  formed. 
However,  if  an  excess  of  sodium  hydroxide  is  used  the  pre- 
cipitate will  dissolve  with  the  formation  of  water  and  sodium 
zincate  according  to  the  equation 

Zn  (OH)2  +  2  NaOH *•  Na2ZnO2  +  2  H2O. 

If  hydrochloric  acid  is  added  to  the  precipitate  of  zinc  hy- 
droxide, the  hydroxide  dissolves  with  the  formation  of  zinc 
chloride  and  water: 

Zn  (OH)2  +  2  HC1 *  ZnCl2  +  2  H2O. 

In  the  reaction  of  both  hydrochloric  acid  and  sodium  hydrox- 
ide on  zinc  hydroxide  water  was  one  of  the  products  of  the 
reaction.  This  shows  that  zinc  hydroxide  must  give  hydroxyl 
ions  to  combine  with  the  hydrogen  ion  of  hydrochloric  acid 
to  form  water,  and  also  that  zinc  hydroxide  must  give 
hydrogen  ions  to  combine  with  the  hydroxyl  ions  of  sodium 
hydroxide  to  form  water.  Zinc  hydroxide  must  then  ionize 
both  as  an  acid  and  as  a  base  : 

2  H+  +  ZnO2—  :<=>:  Zn  (OH)2  ^=±  Zn++  +  2  OH~. 

In  the  presence  of  a  strong  base  zinc  hydroxide  acts  as 
an  acid ;  in  the  presence  of  a  strong  acid  it  acts  as  a  base. 
A  hydroxide  that  ionizes  both  as  an  acid  and  as  a  base  is 
called  an  amphoteric  hydroxide.  Many  of  the  elements  form 
hydroxides  that  show  this  characteristic  to  a  certain  extent. 
The  elements  that  show  this  property  most  prominently  are 
those  elements  that  form  neither  strong  acids  nor  strong 
bases.  Among  these  may  be  mentioned  zinc,  aluminium, 
chromium,  tin,  arsenic,  and  lead.  Further  discussion  will  be 
given  as  these  examples  are  met  with  in  analysis. 


50  NOTES  ON  QUALITATIVE  ANALYSIS 

FORMATION  OF  COMPLEX  IONS 

When  ammonia  gas  (NH3)  is  passed  into  water  it  dissolves, 
partly  by  chemical  union  with  the  water  to  form  ammonium 
hydroxide.  As  a  result  the  following  equilibrium  exists : 

NH8  +  H2O  +=t  NH4OH  T=±  NH4+  +  OH-. 

As  the  solution  contains  ammonia  (NH3)  molecules  as  well 
as  NH4+  and  OH~  ions,  there  is  a  possibility  that  more  than 
one  reaction  will  take  place  when  a  third  substance  is  added. 

(1)  Double  decomposition  may  take  place,  resulting  in  the 
formation  of  the  metallic  hydroxide : 

CrCl8  +  3NH4OH  ^=±  3NH4C1  +  Cr(OH)3. 

(2)  The  metallic  ion  may  unite  with  the  ammonia  (NH3)  to 
form  a  complex  ion.    The  number  of  ammonia  groups  (NH8) 
which  may  unite  with  the  metallic  ion  varies  under  different 
conditio'ns  of  pressure,  temperature,  etc.,  but  it  is  usually 
2,  4,  or  6. 

The  metals  which  form  complex  ions  with  ammonia  are 
silver,  copper,  cadmium,  cobalt,  nickel,  and  zinc.  A  few 
examples  will  illustrate : 

CuSO4  +  nNH4OH  :<=>:  Cu(NH8)nSO4  +  nH2O, 
Cu(OH)2  +  nNH4OH  :<=*:  Cu(NH8)n(OH)2  +  nH2O, 

where  n  represents  some  small  number,  usually  from  2  to  6. 
The  complex  ion  (Cu(NH8)n)  formed  is  positive  and  has  the 
same  valence  as  the  metallic  ion  from  which  it  was  formed. 
The  cyanides  of  the  heavy  metals  dissolve  in  an  excess  of 
the  alkali  cyanides  to  form  complex  compounds,  as  shown  by 
the  following,  where  M  stands  for  the  metal: 

KCu(CN)2,     K2Cd(CN)4,     K8Fe(CN)6,     K4Fe(CN)8, 
(M^(CN),)K2,     (M+  ++(CN)6)K3,     (M++(CN)6)K4.  ' 

Some  of  the  metals  which  form  complex  cyanides  are 
silver,  gold,  copper,  cadmium,  iron,  cobalt,  nickel,  and  zinc. 


COMPLEX  IONS 


51 


A  separation  of  copper  and  cadmium  or  of  cobalt  and  nickel 
may  be  made  by  the  use  of  these  complex  cyanides.  The 
potassium  salts  of  hydroferrocyanic  acid  (K4Fe(CN)6)  and 
of  hydroferricyanic  acid  (K3Fe(CN)6)  are  used  in  making 
tests  for,  or  in  distinguishing  between,  ferric  and  ferrous  iron. 
Certain,  organic  acids  have  the  ability  to  unite  with  some 
metallic  ions  to  form  complex  ions.  The  formula  for  tartaric 
acid  (H2C4H4O6)  is  written  structurally: 

CO  -  OH  (1) 

in  -  OH  (2) 

CH  -  OH  (3) 

io  -  OH  (4) 

in  which  hydrogen  atoms  (1)  and  (4)  split  off  as  ions. 


CO  -  OH  " 

COO    i 

CH-OH 

._^ 

CH-OH 

CH-OH 

CH-OH 

CO  -  OH  , 

.COO 

When  ionized  in  this  manner  such  salts  as  cream  of  tartar 
KH  (C4H4O6)  and  Rochelle  salts  NaK  (C4H4O6)  may  be 
formed.  In  the  presence  of  a  large  excess  of  sodium  or 
potassium  hydroxide  the  hydroxyl  groups  attached  to  the 
CH  groups  may  split  off  hydrogen  ions  which  can  be  replaced 
by  certain  metals.  Nickel  and  copper  ions  under  these  condi- 
tions enter  easily  into  the  tartrate  molecule,  as  illustrated  by 

COONa 
H-0 


CH-0 
COOK 


VV^V^JV 

which  gives  the  ions 

Na+,  K+,  and  C  H  O  Ni~ 

426 


52  NOTES  ON  QUALITATIVE  ANALYSIS 

Such  complex  ions  are  quite  stable,  and  in  the  case  of  the 
nickel  salt  so  few  nickel  ions  are  given  that  no  precipitate 
of  nickel  will  be  produced  in  such  a  solution  by  hydrogen  sul- 
fide.  Advantage  is  sometimes  taken  of  this  in  the  separation 
of  cobalt  from  nickel. 

All  of  these  reactions,  in  which  complex  ions  are  formed, 
are  more  or  less  reversible,  and  hence  there  is  always  present 
in  solution  of  such  salts  a  few  of  the  simple  metallic  ions  as 
well  as  the  complex  ions.  This  is  easily  understood  when  it 
is  remembered  that  sulfide  ion  (S~~~)  precipitates  zinc,  cobalt, 
nickel,  or  copper  as  sulfide  from  solutions  of  their  salts  made 
alkaline  with  ammonium  hydroxide.  This  precipitation  of  the 
metallic  sulfide  could  not  occur  if  all  the  metal  ion  had 
entered  the  ammonia  molecule  to  form  the  complex  ion.  This 
fact  is  important  when  considering  these  complex  ions. 

The  ions  of  the  iron-zinc  group  are  Fe+  +  +  ,  Fe  +  +,  Al  +  +  +, 
Cr+  +  +,  Zn++,  Co  +  +,  Ni  +  +,  Mn  +  +. 

Of  these  metals  aluminium,  chromium,  and  zinc  are  ampho- 
teric,  their  hydroxides  dissolving  in  strong  bases,  and  zinc, 
cobalt,  and  nickel  form  complex  ions  with  ammonia.  Man- 
ganous  hydroxide  is  not  amphoteric,  neither  does  manganese 
form  complex  ions  with  ammonia;  but  being  somewhat  solu- 
ble, manganous  hydroxide  behaves  like  magnesium  hydroxide 
and  dissolves  in  ammonium  chloride  solution,  though  less 
readily  than  the  latter.  Manganous  hydroxide  readily  com- 
bines with  oxygen  from  oxidizing  agents,  or  even  from  the  air, 
going  over  into  the  less  soluble  hydrated  manganese  dioxide 
(MnO(OH)2),  and  in  this  condition  is  not  soluble  in  ammo- 
nium chloride.  Cobalt  and  nickel  sulfides  are  peculiar  in  the 
fact  that  while  they  are  precipitated  only  in  neutral  or  alka- 
line solution,  after  precipitation  these  sulfides  are  not  soluble 
in  dilute  acids.  This  insolubility  is  probably  due  to  a  change 
which  occurs  in  the  sulfides  after  precipitation.  Strong  acids 
or  aqua  regia  dissolve  them. 


THE  IRON-ZINC  GROUP  53 

• 

Ferrous  hydroxide  is  somewhat  similar  to  magnesium  and 
manganous  hydroxide  and  is  not  completely  precipitated  in 
the  presence  of  ammonium  salts.  For  its  complete  separation 
from  manganese  it  is  therefore  necessary  to  oxidize  the  iron 
to  the  ferric  condition  before  precipitation,  as  the  ferric 
hydroxide  is  not  soluble  in  ammonium  salts. 

Chromium  and  manganese  in  their  highest  states  of  oxida- 
tion (chromium  six  and  manganese  six  and  seven)  form 
acids  the  alkali  salts  of  which  are  soluble,  and  whose  ions 
are  characteristically  colored  in  solution.  It  is  thus  possible 
to  identify  these  elements  in  the  presence  of  some  others  of 
this  group.  These  oxidations  may  be  illustrated  by  the 
following  equations,  written  in  steps,  showing  the  oxidation 
of  chromic  hydroxide  to  chromic  acid  by  hydrogen  peroxide 
and  the  oxidation  of  manganous  sulfate  to  permanganic  acid 
by  oxygen  which  may  be  considered  as  obtained  from  any 
sufficiently  strong  oxidizing  agent : 

2Cr(OH)3  — ^Cr,03  +  3H20, 

3H202— ^3H20  +  30, 
Cr203  +  30— ^2Cr03, 
2  Cr08  +  2  H20 ^2  H2CrO4, 

and  2  MnS  O4 >-  2  MnO  +  2  S  O3, 

2  MnO  +  5  O >-  Mn2O7, 

Mn2O7  +  H2O >-  2  HMnO4. 

(2  SO3  +  2  H2O  — >•  2  H2SO4.) 

In  the  first  illustration  the  color  of  the  solution  changes 
from  green  to  yellow,  as  the  chromic  ion  is  green  and  the 
chromate  yellow.  In  the  second  the  manganous  ion  which  is 
almost  colorless  has  been  converted  into  the  intensely  purple 
permanganate  ion.  As  CrO3  is  an  acid  anhydride  only, 
hydrogen  sulfide  will  not  precipitate  chromium  sulfide  nor 
hydroxide  from  solutions  of  chromates.  It  is  therefore  neces- 
sary, if  chromium  might  be  present,  to  reduce  any  chromate 


54  NOTES  ON  QUALITATIVE  ANALYSIS 

% 

to  a  chromic  salt  before  precipitation  of  the  iron-zinc  group 
with  ammonium  sulfide.  The  reduction,  which  must  be 
carried  out  before  iron  is  oxidized,  may  be  accomplished  in 
acid  solution  by  any  good  reducing  agent,  as  hydrogen  sulfide, 
sulfur  dioxide,  alcohol,  etc.  This  is  illustrated  by  the  reduc- 
tion of  potassium  dichromate  to  chromic  sulfate  by  hydrogen 
sulfide  in  sulfuric  acid  solution  : 


2Cr03  -  ^Cr203  +  30, 
3  o  +  3H2S  -  >-  3  H2O  +  3  S  ; 

CrA  +  3  H2S04  —  *  Cr2(s°4)8  +  3  H2°  ; 

K20  +  H2S04  —  *  K2S04  +  H20. 

The  color  of  the  solution  changes  from  the  orange  of  the 
dichromate  ion  to  the  green  of  the  chromic  ion. 

Most  of  the  metals  of  this  group  when  fused  with  borax 
impart  a  characteristic  color  to  the  borax  glass  so  formed. 
In  many  cases  these  colored  glasses  furnish  satisfactory  tests 
for  the  elements.  Borax  (Na2B4O7)  may  be  considered  as 
made  up  from  two  molecules  of  sodium  meta-borate  and  one 
molecule  of  boric  anhydride  thus  : 

2NaB0.2.B203. 

Like   other  acid  anhydrides,   boric   anhydride  is  capable  of 
combining  with  metallic  oxides  to  form  salts,  as 

CuO  +  B203  -  *•  Cu(B02)2. 

The   reaction  of  cobalt   in   the  borax  bead   might  then   be 
written  : 

2  NaB02  •  B208  +  CoO  --  H  2  NaBO2  .  Co(BO2)2. 
This  compound  is  intensely  blue. 


EXPERIMENTS  OF  THE  IRON-ZINC  GROUP      55 

SEPARATION  AND  IDENTIFICATION  OF   THE  MEMBERS  OF 
THE  IRON-ZINC  GROUP 

Before  beginning  the  following  experiments  review  the 
results  obtained  in  Experiments  12  and  13,  pages  47~48. 

Experiment  14 

1.  To  5  cc.  of  ferric  chloride  solution  add  a  drop  or  two  of  am- 
monium hydroxide,  shake,  and  then  add  an  excess  of  this  reagent 
(see  page  50).    Now  add  2  cc.  of  ammonium  chloride  solution  and 
observe  if  any  change  occurs. 

2.  Repeat  1,  using  solutions  of  potassium  alum,  chrome  alum, 
cobalt  chloride,  nickel  chloride,  zinc  sulfate,  and  manganese  sulfate 
instead  of  ferric  chloride. 

3.  Repeat  1  and  2,  using  sodium  hydroxide  in  place  of  am- 
monium hydroxide  and  omitting  the  ammonium  chloride.    Note 
how  the  results  in  3  differ  from  those  obtained  in  1  and  2  and 
explain. 

Experiment  15 

Add  sodium  hydroxide  to  a  chrome  alum  solution  until  the  pre- 
cipitate first  formed  is  dissolved.  Now  add  about  1  gram  of  sodium 
peroxide  and  boil  until  all  the  peroxide  is  decomposed.  Notice, 
and  account  for,  the  change  in  color.  Acidify  one  half  of  this  solu- 
tion with  sulfuric  acid  and  introduce  hydrogen  sulfide  until  the 
color  is  changed.  Acidify  the  other  half  of  the  solution  with  acetic 
acid  and  divide  it  into  two  parts.  To  one  part  add  barium  chloride 
solution  and  to  the  other  lead  acetate.  Explain  all  reactions. 

Experiment  16 

1.  Prepare  some  aluminium  hydroxide  and  fuse  a  portion  of  it 
with  sodium  carbonate  on  a  piece  of  platinum  foil  (or  a  piece  of 
charcoal  may  be  substituted  for  the  platinum)  which  is  held  in 
the  oxidizing  portion  of  a  blowpipe  flame.   Moisten  the  white  fused 
mass  with  one  drop  of  cobalt  nitrate  solution  and  fuse  again. 

2.  Repeat  1,  using  any  zinc  salt  instead  of  the  aluminium  hy- 
droxide.   Compare  the  two  results. 


56  NOTES  ON  QUALITATIVE  ANALYSIS 

Experiment  17 

1.  Take   three   portions  each  of  ferric  chloride  solution  and 
freshly  prepared  ferrous  ammonium  sulfate  solution.  To  one  por- 
tion of  each  add  potassium  ferrocyanide  solution;  to  a  second, 
potassium  ferricyanide  solution ;  and  to  the  third,  potassium  thio- 
cyanate  solution.    Note  carefully  the  colors  in  each  case. 

2.  Repeat  1,  using  three  portions  of  a  ferric  chloride  solution 
which  has  been  acidified  with  hydrochloric  acid  and  has  had 
hydrogen  sulfide  passed  through  it  for  a  few  minutes. 

3.  Repeat  1,  using  a  solution  of  ferrous  ammonium  sulfate 
which  has  been  boiled  for  a  few  minutes  with  1  cc.  of  concentrated 
nitric  acid.   Explain  the  results. 

SCHEME  FOR  ANALYSIS  OF  THE  IRON-ZINC  GROUP  * 


Besides  the  metallic  ions  of  this  group  the  solution  may  contain 
the  metallic  ions  of  the  alkaline  earth  and  alkali  groups. 


Warm  the  solution,  add  NH4C1  and  NH4OII  until,  after  shaking, 
the  solution  smells  strongly  of  ammonia.  Keep  near  the  boiling 
point  and  saturate  with  H2S.  Filter;  wash  as  quickly  as  possible 
with  water  containing  a  little  (NH4)2S  to  prevent  oxidation.  Test 
filtrate  with  (NH4)2S  to  make  sure  all  of  group  is  precipitated.  If 
the  filtrate  is  brown,  an  excess  of  (XH4)2S  has  been  used  and  some 
nickel  sulfide  has  dissolved.  In  this  case  boil  the  brown  filtrate 
with  acetic  acid  and  filter  through  a  fresh  filter. 


Filtrate  contains  metals  of  alkaline  earth  and  alkali  groups  and  after 
boiling  off  H2S  is  reserved  to  be  tested  for  these  groups  (see  page  38). 


Precipitate :  A1(OH)3,  Cr(OH)3,  Fe2S3,  FeS,  MnS,  ZnS,  CoS,  NiS 
Remove  precipitate  from  the  filter  and  treat  with  15-25  cc.  of 
dilute  HC1  (dilute  acid  of  the  laboratory  and  an  equal  volume  of 
water),  grinding  the  precipitate  with  a  test  tube  as  a  pestle.  Do  not 
warm.  The  treatment  should  not  be  longer  than  five  minutes  before 
filtering  off  the  solution.  Wash  any  black  residue,  f 


*  If  phosphates  or  oxalates  may  be  present  (see  page  72). 
t  This  scheme  for  analysis  is  continued  on  opposite  page. 


ANALYSIS  OF  THE  IRON-ZINC  GROUP 


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68  NOTES  ON  QUALITATIVE  ANALYSIS 

Experiment  18 

To  two  or  three  drops  of  manganese  sulfate  solution  add  5  cc. 
of  dilute  sulfuric  acid  and  enough  lead  dioxide  to  fill  the  rounded 
bottom  of  the  test  tube.  Boil  for  a  minute  or  two  and  allow  the 
suspended  material  to  settle.  Note  the  color  of  the  solution  and 
explain.  Chlorides  must  not  be  present  in  making  this  test  for 
manganese.  Why  ? 

Experiment  19 

1.  Bend  a  platinum  wire  around  the  point  of  a  lead  pencil  so 
as  to  make  a  small  loop.    Heat  the  loop  to  redness,  dip  it  into  a 
little  borax,  and  then  fuse  the  portion  sticking  to  the  wire  into  a 
clear  bead.   While  the  bead  is  still  hot  touch  it  to  a  very  small 
particle  of  cobalt  salt  and  then  heat  strongly  in  the  oxidizing 
flame.    Cool,  and  note  the  color  of  the  bead.   Kepeat  this  experi- 
ment, using  a  nickel  salt  and  a  manganese  salt  instead  of  the  cobalt. 

2.  Make  a  bead  as  described  in  1,  using  sodium  carbonate  to 
which  is  added  a  small  particle  of  potassium  nitrate.   Touch  the 
bead  to  a  small  particle  of  manganese  salt  and  fuse  in  the  oxidiz- 
ing flame.    Cool,  and  note  the  color. 

Experiment  20* 

Precipitate  a  little  nickel  sulfide,  filter,  and  dissolve  the  pre- 
cipitate in  a  few  drops  of  aqua  regia.  Evaporate  almost  to  dryness, 
stopping  before  the  last  drop  has  disappeared  in  order  to  avoid 
overheating.  Add  a  few  cubic  centimeters  of  water,  then  a  large 
excess  of  solid  sodium  acetate  and  a  few  drops  of  dimethyl 
glyoxime  reagent. 

Experiment  21 

To  a  solution  of  zinc  sulfate  add  sodium  hydroxide  until  the 
precipitate  first  formed  dissolves.  Divide  this  solution  into  two 
parts.  Pass  hydrogen  sulfide  into  one  part.  Acidify  the  other 
part  with  acetic  acid  and  pass  in  hydrogen  sulfide.  Explain. 

*  Mix  together  one  volume  of  a  cobalt  chloride  solution  with  one-half 
volume  of  nickel  chloride.  Add  to  this  mixture  an  excess  of  sodium  potas- 
sium tartrate  and  of  sodium  hydroxide.  Saturate  the  solution  completely 
with  hydrogen  sulfide.  Filter,  and  note  the  very  dark-brown-colored  filtrate. 
What  is  the  precipitate  ?  What  causes  the  color  in  the  filtrate  ? 


THE  COPPER  GROUP  59 

THE  COPPER  GROUP 

The  ions  to  be  tested  for  in  this  group  are  Hg++,  Pb++, 
Cu++,  Bi+++,  and  Cd++. 

Mercury,  copper,  and  cadmium  are  marked  by  their  tend- 
ency to  form  complex  ions  with  ammonia  or  potassium  cyanide 
(see  page  50),  while  lead  is  amphoteric  (see  page  49),  forming 
both  lead  salts  and  plumbites,  the  alkali  salt  of  the  latter  being 
soluble.  Lead  also  is  interesting  in  that  it  forms  an  insoluble 
sulfate.  Mercury  and  cadmium  are  unusual  in  that  their 
salts,  particularly  the  chlorides,  are  very  slightly 'ionized,  and 
this  ionization  may  be  still  further  decreased  by  the  addition 
of  a  common  negative  ion.  Advantage  is  sometimes  taken  of 
this  in  the  separation  of  cadmium  from  copper.  Copper  forms 
two  series  of  compounds,  cuprous  (Cu+)  and  cupric  (Cu++), 
but  the  former  are  so  easily  oxidized  to  the  latter  that  they 
are  never  used  for  identification.  Most  cupric  salts  are  highly 
colored  and  can  be  easily  identified  by  this  means.  Like 
copper,  mercury  forms  two  series  of  compounds,  mercurous 
(Hg+)  and  mercuric  (Hg+  +).  The  mercurous  ion,  forming  an 
insoluble  chloride,  is  always  tested  for  in  the  silver  group. 
As  mercurous  salts  oxidize  easily  to  mercuric,  the  latter  will 
nearly  always  be  present  if  the  mercurous  ion  has  been 
found  in  the  silver  group.  Mercuric  salts  are  rather  easily 
reduced  to  mercurous  or  even  to  free  mercury  by  strong 
reducing  agents,  such  as  stannous  chloride,  and  this  property 
is  often  used  for  their  identification. 

Bismuth,  although  it  occurs  in  the  fifth  group  of  the 
periodic  table,  behaves  mostly  as  a  metal  and  shows  few 
unusual  reactions,  except  that  its  salts  are  considerably  hydro- 
lyzed.  The  products  of  hydrolysis  are  mostly  insoluble  basic 
salts,  one  of  the  least  soluble  and  most  characteristic  being 
the  oxychloride  (BiOCl).  The  reaction  is  easily  reversed  and 
the  salt  dissolves  again  in  an  excess  of  acid. 


60  NOTES  ON  QUALITATIVE  ANALYSIS 

Experiment  22 

1.  Acidify  a  solution  of  mercuric  chloride  with  dilute  hydro- 
chloric acid,  heat  to  boiling,  and  saturate  with  hydrogen  sulfide. 
Filter,  and  test  the  solubility  of  the  precipitate  in  boiling  dilute 
nitric  acid.   Keep  the  volume  of  acid  constant  during  boiling. 

2.  Repeat  1,  using  3-5  cc.  of  each  of  the  following  solutions : 
lead  acetate,  bismuth  chloride,  copper  sulfate,  and  cadmium  sulfate. 

Experiment  23 

1.  Using  solutions  of  lead  acetate,  bismuth  chloride,  copper 
sulfate,  and  cadmium  sulfate,  add  to  each  one  drop  of  ammonium 
hydroxide,  shake,  and  then  add  an  excess. 

2.  Filter  off  the   precipitate  formed  in  the  bismuth  chloride 
and  wash  the   precipitate  into  the  tip  of  the  filter  with  a  little 
warm  water.    Now  pour  a  few  drops  of  dilute  hydrochloric  acid 
on  the  precipitate  in  the  filter  and  allow  one  to  two  drops  to 
fall  into  a  beaker  containing  at  least  100  cc.  of  cold  water. 

Experiment  24 

Add  dilute  sulfuric  acid  to  a  solution  of  lead  acetate.  Filter, 
and  wash  the  precipitate.  Dissolve  in  sodium  hydroxide,  acidify 
the  solution  with  acetic  acid,  and  then  add  potassium  dichromate. 

Experiment  25 

Mix  together  a  few  cubic  centimeters  of  copper  sulfate  and 
cadmium  sulfate.  Acidify  3  cc.  of  the  mixture  with  hydrochloric 
acid  and  add  solid  sodium  chloride  until  the  solution  is  saturated. 
Warm  the  solution  and  saturate  it  with  hydrogen  sulfide.  Filter 
through  a  dry  filter  paper.  Dilute  the  filtrate  with  several  times 
its  volume  of  water  and  again  saturate  it  with  hydrogen  sulfide. 

Experiment  26 

Acidify  a  solution  of  copper  sulfate  with  5  cc.  of  concentrated 
nitric  acid.  Warm  and  pass  in  hydrogen  sulfide  for  two  or  three 
minutes.  Recall  that  HN03  is  a  strong  oxidizing  agent  and  ac- 
count for  the  brown  vapors  and  the  yellowish  precipitate  which 
forms.  Nitric  acid  should  always  be  avoided  when  precipitating 
with  hydrogen  sulfide.  How  could  it  be  removed  if  present  ? 


ANALYSIS  OF  THE  COPPER  GROUP 


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62  NOTES  ON  QUALITATIVE  ANALYSIS 

THE  AKSENIC  GROUP 

The  ions  to  be  tested  for  in  this  group  are  As+++,  As+++++, 
Sb+++,  Sb+++++,  Sn++,  and  Sn++++. 

The  fact  has  already  been  mentioned  that  arsenic,  anti- 
mony, and  tin  are  decidedly  nonmetallic,  particularly  in  their 
higher  valences,  and  that  this  property  is  taken  advantage 
of  in  separating  them  from  the  metals  of  the  copper  group. 

Arsenic,  in  its  highest  valence,  forms  a  rather  strong  acid, 
arsenic  acid  (HgAsO4),  similar  to  phosphoric  acid,  and  any 
basic  properties  which  it  may  show  are  very  weak.  For  this 
reason  it  is  usually  very  difficult  to  precipitate  arsenic  from 
arsenic  solutions  by  means  of  hydrogen  sulfide.  Arsenious 
acid  is  a  weaker  acid  and  shows  more  pronounced  basic 
properties,  so  that  arsenic  trisulfide  may  be  precipitated 
without  difficulty  from  such  solutions. 

Hydrogen  sulfide  slowly  reduces  arsenic  acid  to  arsenious 
acid,  which  is  then  precipitated  as  the  trisulfide  along  with 
some  pentasulfide.  Heat  increases  the  rate  of  this  reduction, 
and  a  large  excess  of  hydrochloric  acid  favors  the  precipita- 
tion of  the  pentasulfide. 

Arsenic  acid,  being  very  similar  to  phosphoric  acid,  easily 
forms  such  compounds  as  magnesium  ammonium  arsenate 
(MgNH4AsO4),  this  compound  forming  a  characteristic,  crys- 
talline precipitate  in  strong  ammoniacal  solution. 

Antimonic  acid  shows  more  basic  properties  than  arsenic 
acid,  and  the  pentasulfide  is  easily  precipitated. 

Antimony  compounds  are  easily  reduced  by  hydrogen,  in 
the  presence  of  metals,  to  metallic  antimony,  which  appears 
as  a  fine  black  deposit. 

Both  stannous  and  stannic  acids  show  some  basic  proper- 
ties, hence  stannous  and  stannic  sulfides  are  precipitated 
without  difficulty.  However,  these  sulfides  are  soluble  in 
much  hydrochloric  acid  and  can  only  be  precipitated  when 


THE  ARSENIC  GKOUP  63 

the  concentration  of  this  acid  is  small.  Hence  the  necessity, 
frequently,  of  greatly  diluting  the  solution  after  arsenic  has 
been  precipitated.  This  difference  in  the  solubility  of  the 
sulfides  of  arsenic,  antimony,  and  tin  in  hydrochloric  acid  is 
often  taken  advantage  of  in  separating  them. 

Stannous  salts  are  easily  oxidized  to  stannic,  so  that  stan- 
nous  chloride  readily  reduces  ferric  salts  to  ferrous,  mercuric 
to  mercurous  or  free  mercury,  and  molybdic  acid  to  lower 
oxides  of  molybdenum. 

SCHEME  FOR  ANALYSIS  OF  THE  ARSENIC  GROUP 


The  solution  containing  compounds  of  arsenic,  antimony,  and  tin 
is  acidified  with  about  2  cc.  of  concentrated  HC1  warmed  and  satu- 
rated with  H2S.  Now  dilute  until  the  acidity  is  .25  molar  and  again 
saturate  with  H2S.  Filter,  and  wash  the  precipitate  with  hot  water. 
Transfer  the  precipitated  sulfides  to  an  evaporating  dish  and  add  to 
them  3  or  4  cc.  of  concentrated  hydrochloric  acid.  Care  must  be 
taken  not  to  dilute  the  acid  in  any  way.  Grind  up  the  precipitate 
with  the  acid  and  warm  the  mixture  as  long  as  there  is  a  brisk 
evolution  of  hydrogen  sulfide.  Cool  and  filter  through  a  dry  filter. 
Saturate  the  filtrate  with  hydrogen  sulfide.  If  a  yellow  precipitate 
of  arsenic  sulfide  forms,  it  must  be  filtered  out  of  the  solution. 


Residue :  As2S3,  As2S5 
Dissolve  the  precipitate 
in  concentrated  HC1  con- 
taining a  few  crystals  of 
potassium  chlorate  and 
test  with  NH4OH  and 
MgCl2  (see  Experiment 
28,  p.  64). 


Filtrate :  SbCl3,  SnCl4,  SnCl2 
Boil  until  the  hydrogen  sulfide  is  en- 
tirely removed  from  the  filtrate,  which 
should  then  have  a  volume  of  2  cc. 
Dilute  with  an  equal  volume  of  water 
and  test  for  antimony  and  tin  as  de- 
scribed in  Experiment  29,  p.  64,  by 
adding  nails  etc. 


64  NOTES  ON  QUALITATIVE  ANALYSIS 

Experiment  27 

Take  solutions  of  sodium  arsenite,  of  antimony  chloride,  and 
of  stannous  chloride  and  add  to  each  about  1  cc.  of  dilute  hydro- 
chloric acid.  Warm  each  solution  and  saturate  with  hydrogen 
sulfide.  Test  the  solubilty  of  each  precipitate  in  concentrated 
hydrochloric  acid.  To  the  solutions  of  the  precipitates  which 
have  dissolved  add  water,  drop  by  drop,  until  the  precipitate  re- 
appears. Note  approximately  the  amount  of  water  added  before 
each  precipitate  reappears. 

Experiment  28 

Precipitate  some  arsenic  sulfide  and  dissolve  it  in  concentrated 
hydrochloric  acid  to  which  a  few  crystals  of  potassium  chlorate 
have  been  added.  Evaporate  nearly  to  dryness,  add  a  small  amount 
of  water,  an  excess  of  ammonium  hydroxide,  and  some  magnesium 
chloride  solution.  The  precipitate  may  not  form  immediately.  If 
not,  shake  vigorously  and  let  stand. 

Experiment  29 

Mix  together  a  few  cubic  centimeters  of  antimony  chloride  and 
stannous  chloride.  Add  a  few  drops  of  HC1  and  several  small 
iron  tacks  or  nails  to  the  solution  and  heat  to  boiling  for  two  or 
three  minutes.  A  vigorous  evolution  of  hydrogen  will  occur  and 
a  dark  granular  deposit  appear.  Filter,  and  treat  the  residue  and 
filtrate  as  follows  :  Dissolve  the  residue  in  about  5  cc.  of  hot  tar- 
taric  acid  solution  containing  three  or  four  drops  of  dilute  nitric 
acid.  Place  in  a  precipitation  flask  and  pass  in  hydrogen  sulfide, 
preferably  keeping  the  delivery  tube  just  above  the  liquid.  Shake 
the  solution.  Divide  the  filtrate  from  the  granular  residue  into 
two  parts.  To  one  part  add  a  little  mercuric  chloride  solution. 
Dilute  the  other  part  with  30-50  volumes  of  water  and  add  2-3  cc. 
of  a  neutral  ammonium  molybdate  solution. 


SEPARATION  OF  ARSENIC  AND  COPPER  GROUPS    65 


SCHEME  FOR  SEPARATION  OF  THE  ARSENIC  AND 
COPPER  GROUPS 


The  solution,  which  may  contain  metals  of  both  groups,  should  be 
clear.  It  is  acidified  rather  strongly  with  HC1,  warmed,  then  saturated 
with  H2S  as  long  as  any  precipitate  forms.  Filter,  dilute  the  filtrate 
until  the  acidity  is  approximately  .25  molar,  arid  again  saturate  with 
H2S.  Filter  through  the  same  filter  and  test  the  filtrate  by  passing 
in  more  H2S.  Wash  the  precipitate  two  or  three  times  with  hot  water. 


Precipitate :  HgS,  PbS,  Bi2S3,  CuS,  CdS,  As2S3, 

As2S5,  Sb2S3,  Sb2S5,  SnS,  SnS2 
Transfer  a  small  portion  of  the  precipitate  to  a 
test  tube,  add  1  cc.  (NH4)2S.C,  and  warm  to  about  60°. 
Filter,  and  test  the  filtrate  for  members  of  arsenic 
group  by  acidifying  with  HC1.  If  only  a  white 
or  yellowish  precipitate  of  sulfur  is  thrown  down, 
the  arsenic  group  is  not  present,  but  if  a  yellow, 
orange,  or  brown  flocculent  precipitate  is  obtained, 
the  arsenic  group  is  present.  If  the  arsenic  group 
is  present,  transfer  the  whole  precipitate  obtained 
by  H2S  to  a  beaker  or  evaporating  dish  and  treat 
with  (NH4)2SX,  as  directed  above.  Filter.  Wash. 


Filtrate 
Contains  iron- 
zinc  and  other 
groups,  and  is  set 
aside  to  be  tested 
for  them  later 
(see  page  56). 


Filtrate:  (NH4)3AsS4,  (NH4)3SbS4, 

(NH4)2SnS3 

Acidify  filtrate.  Filter,  wash,  transfer 
to  an  evaporating  dish  and  analyze  the 
precipitate  according  to  scheme  for  anal- 
ysis of  arsenic  group,  beginning  where 
the  precipitate  is  ground  with  3  or  4  cc. 
of  HC1  (p.  63). 


Residue 

Sulfides  of  the  copper 
group.  To  be  analyzed 
according  to  scheme  for 
analysis  of  copper  group, 
beginning  by  dissolving 
the  precipitate  in  dilute 
HNO3  (p.  61). 


66  NOTES  ON  QUALITATIVE  ANALYSIS 

THE  SILVER  GROUP 

This  group  includes  silver,  lead,  and  mercurous  mercury. 
Silver  forms  but  one  series  of  salts,  and  in  this  it  is  univa- 
lent.  Mercury  forms  two  series  of  salts.  In  the  mercuric 
salts  it  has  a  valence  of  two,  while  in  the  mercurous  salts  it 
has  a  valence  of  one.  Lead  most  commonly  shows  a  valence 
of  two.  It  is,  however,  tetravalent  in  some  of  its  compounds, 
and  also  occurs  in  the  acid  ions  PbO8  ,  PbO4~"  "."• 

Silver  and  mercury  have  a  marked  tendency  to  enter  into 
complex  ions,  as  KAg(CN)2,  Ag(NH3)2Cl,  NH2HgCl,  while 
lead  behaves  in  a  more  normal  manner.  The  formation  of 
these  complex  ions  plays  an  important  part  in  the  separation 
and  identification  of  these  elements. 

The  chlorides  of  silver  and  mercurous  mercury  are  very 
difficultly  soluble  in  water.  Lead  chloride  is  only  sparingly 
soluble  in  cold  water,  but  is  soluble  in  hot  water.  One  hun- 
dred cubic  centimeters  of  water  will  dissolve  about  1.1  grams 
of  PbCl2,  0.000038  grams  of  HgCl,  and  0.00015  grams  of 
AgCl  at  ordinary  temperatures.  At  100°  the  same  amount 
of  water  will  dissolve  about  3.5  grams  of  PbCl2. 

As  a  group  precipitant  hydrochloric  acid  will  be  used. 
This  will  precipitate  silver  and  mercurous  mercury  almost 
completely,  but  will  leave  some  of  the  lead  in  solution.  If 
the  solution  is  very  dilute  with  respect  to  lead,  no  precipitate 
of  lead  may  be  thrown  down  by  hydrochloric  acid,  but  will 
be  precipitated  with  the  copper-arsenic  groups  as  sulfide  and 
can  be  identified  there.  Lead  should  always  be  tested  for  in 
the  copper  group,  whether  it  is  found  in  the  silver  group  or 
not.  Mercuric  mercury  will  be  precipitated  as  sulfide  with 
the  copper  arsenic  groups. 

Silver  chloride  dissolves  in  ammonium  hydroxide,  forming 
the  soluble  complex  salt  Ag(NH3)2Cl.  This  compound  gives 
so  few  silver  ions  that  a  chloride  will  cause  no  precipitate,  in 


THE  SILVER  GROUP  67 

neutral  or  alkaline  solution.  If  the  solution  is  acidified  the 
complex  ammonia  ion  will  be  broken  up  and  the  silver 
chloride  reprecipitated. 

Mercuric  and  mercurous  chlorides  interact  with  ammonium 
hydroxide  to  form  the  white  insoluble  compound  HgNH2Cl. 

HgCl2  +  2  NH4OH  — *•  HgNH2Cl  +  NH4C1  +  2H2O. 

2  HgCl  +  2  NH4OH  — *•  HgNH2Cl  +  Hg  +  NH4C1  +  2  H2O. 

The  mercurous  compound  gives  a  black  coloration,  due  to 
the  formation  of  metallic  mercury. 

Lead  forms  an  insoluble  chromate  PbCrO4,  which  is  solu- 
ble in  strong  alkalies,  with  the  formation  of  a  compound 
with  the  PbO2  ion.  This  shows  the  amphoteric  character  of 
lead  hydroxide. 

Bromine  acts  as  an  oxidizing  agent,  oxidizing  mercurous 
compounds  to  mercuric  compounds. 

As  mercurous  compounds  are  so  easily  oxidized  to  mercu- 
ric compounds,  if  the  former  are  found  the  latter  will  nearly 
always  be  present,  and  will  be  identified  in  the  copper  group. 

Experiment  30 

Use  solutions  of  the  nitrates  of  silver,  lead,  and  mercurous 
mercury.  To  each  solution  add  hydrochloric  acid.  Bring  the 
precipitated  chlorides  on  filters,  and  wash  once  with  cold  water. 
Treat  each  precipitate  with  about  50  cc.  of  hot  water  in  portions 
of  about  10  cc.  at  a  time.  Divide  the  solution  from  the  precipitate 
which  dissolves  into  two  parts.  To  one  part  add  potassium 
dichromate  solution  and  to  the  other  dilute  sulfuric  acid.  Add 
sodium  hydroxide  to  the  precipitated  chromate  and  then  acidify 
with  acetic  acid. 

To  the  two  precipitates  which  did  not  dissolve  in  hot  water 
add  1  cc.  of  ammonium  hydroxide,  catching  what  runs  through  in 
clean  test  tubes.  Make  the  solution  in  each  test  tube  acid  with 
dilute  nitric  acid.  Treat  the  black  residue  with  a  saturated 
solution  of  bromine  water. 


68 


NOTES  ON  QUALITATIVE  ANALYSIS 


Experiment  31 

Add  stannous  chloride  to  solutions  of  mercuric  and  mercurous 
salts.  Stannous  chloride  acts  as  a  reducing  agent,  being  itself 
oxidized  to  stannic  compounds. 

Experiment  32 

Heat  silver  chloride  with  sodium  carbonate  on  a  piece  of 
charcoal  with  the  blowpipe.  Repeat  with  lead  and  mercury  salts. 

Experiment  33 

Expose  silver  chloride  to  sunlight  for  a  short  time. 

SCHEME  FOR  ANALYSIS  OF  THE  SILVER  GROUP 


Solution  contains  silver,  lead,  and  mercurous  mercury.  Acidify 
with  HNO8.  Add  HC1  until  precipitation  is  complete.  Filter,  and 
wash  precipitate  with  cold  water. 


Precipitate  :  AgCl,  PbCl2,  HgCl 

Wash  the  precipitate  on  the" filter  with  5  cc.  of  hot  water,  catching 
the  water  in  a  test  tube,  and  divide  into  two  parts.  Test  for  lead  by 
adding  to  one  part  a  few  drops  of  H2SO4  and  to  the  other  K2Cr2O7 
solution.  If  lead  is  found,  continue  washing  with  hot  water  until 
the  last  portion  of  the  wash  water  gives  no  precipitate  with  H2SO4. 


'  Residue :  AgCl,  HgCl 
Treat  with  warm  ammonium  hydroxide. 


Black  Residue : 
HgNH2Cl  +  Hg 
Shows  the  pres- 
ence of  mercury. 


Filtrate:  Ag(NH3)2Cl 
Acidify  with  HNO3. 
White  curdy  precipi- 
tate shows  the  pres- 
ence of  silver. 


Filtrate:  PbCl2 
Add  acetic  acid  and  po- 
tassium chromate.  A  yel- 
low precipitate  of  PbCrO4 
shows  the  presence  of  lead. 
This  may  be  confirmed 
by  dissolving  in  sodium 
hydroxide  and  reprecipi- 
tating  with  acetic  acid. 


If  a  large  amount  of  mercurous  chloride  and  a  small  amount  of 
silver  chloride  is  present,  ammonium  hydroxide  may  fail  to  dissolve 
the  silver  chloride.  If  no  test  for  silver  is  obtained  and  mercury  is 
present,  wash  the  black  residue  repeatedly  with  hydrochloric  acid 
and  saturated  bromine  water.  Wash  the  residue  with  hot  water  and 
repeat  treatment  with  ammonium  hydroxide  and  nitric  acid. 


THE  ACID  IONS  69 

ANALYSIS  FOE  ACID  IONS 

In  the  systematic  analysis  for  metal  ions  these  ions  are 
divided  into  very  definite  groups,  and  in  general  the  ions  of 
any  one  group  are  tested  for  in  the  absence  of  all  other  groups. 

Acid  ions  also  are  usually  divided  into  groups,  but  this 
grouping  shows  considerable  variation,  depending  upon  the 
method  of  analysis  used.  The  grouping  generally  depends 
upon  the  behavior  of  the  ions  with  certain  reagents,  such  as 
barium  chloride  and  silver  nitrate,  either  in  acid  or  neutral 
solution.  As  a  rule  the  groups  are  not  separated  from  each 
other  for  analysis.  In  an  analysis  for  the  acid  ions  two  gen- 
eral plans  may  be  followed.  In  one  the  metal  ions,  other 
than  those  of  the  alkali  metals,  are  removed  before  testing 
for  the  acid  ions,  and  the  tests  frequently  made  in  neutral 
or  alkaline  solution.  If  these  ions  were  not  removed  many  of 
the  tests  would  be  interfered  with  by  them.  For  example, 
in  testing  for  a  nitrate,  if  barium  were  present  the  addition  of 
ferrous  sulfate  would  produce  a  precipitate  of  barium  sulfate 
which  would  interfere  more  or  less  with  the  test.  In  the 
other  plan  the  acid  ions  are  tested  for  in  acid  solution,  and 
metal  ions  as  a  rule  do  not  interfere.  In  this  outline  so  few 
acid  ions  will  be  studied  that  no  grouping  will  be  attempted, 
but  each  acid  will  be  identified  by  a  special  test.  The  method 
in  which  the  metal  ions  other  than  the  alkali  metal  ions  are 
removed  will  be  used. 

It  must  be  remembered  that  usually  few  acid  ions  will 
be  found  in  any  mixture.  Much  information  can  often  be 
gained  as  to  the  acid  ions  which  may  be  present  or  absent 
in  a  mixture  or  compound  by  observing  the  solubility  of  the 
substance  in  water  and  acids  when  the  nature  of  the  metal 
ions  is  known.  For  example,  the  sulfate  ion  and  barium  ion 
could  not  both  be  present  in  an  unknown  which  is  soluble 
in  water  or  dilute  acids. 


70  NOTES  ON  QUALITATIVE  ANALYSIS 

The  acid  ions  to  be  studied  in  this  outline  are  chloride 
(C1-),  nitrate  (NO,-),  sulfate  (SO4~),  phosphate  (PO4 — ), 
carbonate  (CO,—),  acetate  (C2H3O2~),  tartrate  (C4H4O6~ ), 
and  borate  (BO3  ). 

PREPARATION  OF  THE  SOLUTION 

If  the  unknown  contains  cations  higher  than  the  alkali 
group,  these  must  be  removed  before  testing  for  anions.  To 
do  this  mix  some  of  the  original  with  water  (complete  solution 
is  not  necessary),  boil,  and  to  the  hot  solution  add  saturated 
Na2CO8  solution  so  long  as  a  precipitate  is  formed.  Boil  for 
a  few  minutes.  This  precipitates  as  carbonates  the  metals 
which  might  interfere.  Filter  and  test  for  NO8~,  Cl~,  SO4  , 
and  PO4 ions  in  separate  portions  of  this  solution.  Each 
portion  to  be  used  in  making  a  test  must  be  acidified  as  directed 
in  the  preliminary  experiments.  The  CO3  ,  C2H3O2~,  and 
C4H4O6  ions  may  be  tested  for  in  separate  portions  of  the 
original  unknown. 

If  phosphate  or  tartrate  ions  are  found,  these  must  be  re-, 
moved  before  testing' for  certain  of  the  cations.    For  special 
separation   in   case   of   the  presence   of   phosphate  ion,   see 
page  72. 

In  case  tartrate  is  found,  this  is  decomposed  before  testing 
for  any  cations  by  treating  the  original  with  concentrated 
H2SO4  and  evaporating  to  dryness.  Repeat  until  there  is  no 
odor  of  tartrate.  The  residue  is  then  used  for  tests  for 
cations  in  the  usual  way. 


THE  ACID  IONS  71 

TESTS  FOR  ANIONS  (SO-CALLED  ACIDS) 
PRELIMINARY  TESTS 

Chloride  (Cl~).  Add  to  a  solution  of  hydrochloric  acid  a  few 
drops  of  a  silver  nitrate  solution.  Test  the  solubility  of  the  pre- 
cipitate in  dilute  nitric  acid,  and  in  ammonium  hydroxide.  After 
dissolving  the  precipitate  in  ammonium  hydroxide  acidify  the 
solution  with  nitric  acid. 

Nitrate  (N03~).  Add  to  a  solution  of  potassium  or  sodium  nitrate 
about  one  half  its  volume  of  freshly  prepared  ferrous  sulfate  solu- 
tion, which  has  been  acidified  with  dilute  sulfuric  acid,  and  mix 
the  two  solutions.  Incline  the  test  tube  and  carefully  pour  down 
the  side  of  the  tube  a  small  quantity  of  concer^trated  sulfuric  acid, 
being  careful  not  to  mix  the  two  liquids  more  than  necessary. 
The  mixture  must  be  kept  cool.  What  forms  at  the  juncture  of 
the  two  liquids  ? 

Sulfate  (S04~  ~).  To  a  solution  of  potassium  or  sodium  sulfate, 
previously  acidified  with  HN03  or  HC1,  add  barium  chloride 
solution. 

Phosphate  (P04  ).  Acidify  a  few  drops  of  a  solution  of  a 
soluble  phosphate  with  nitric  acid.  Add  an  excess  of  ammonium 
molybdate  solution,  and  warm. 

Carbonate  (C03 — ).  Carbonates  when  treated  with  acids  effer- 
vesce, due  to  the  evolution  of  carbon  dioxide,  a  colorless,  odorless, 
noncombustible  gas.  Treat  a  carbonate  with  a  dilute  acid  and 
pass  the  gas  formed  into  lime  water. 

Acetate  (C2H302~).  Add  to  sodium  acetate  a  little  sulfuric  acid 
(1:1),  warm,  and  note  the  odor.  To  another  portion  add  alcohol 
and  sulfuric  acid,  warm,  and  note  odor.  The  odor  in  the  first 
case  is  due  to  acetic  acid,  and  in  the  latter  to  ethyl  acetate 
(CH3COOC2H6). 

Tartrate  (C4H406 — ).  Heat  a  small  amount  of  solid  sodium  tar- 
trate  or  sodium  potassium  tartrate  on  a  crucible  lid.  Note  the 
odor  of  burnt  sugar.  The  test  is  made  more  conclusive  by  first 
moistening  the  tartrate  with  concentrated  sulfuric  acid  and  then 
heating. 


72  NOTES  ON  QUALITATIVE  ANALYSIS 

Borate  (B03 ).    Evaporate  just  to  dryness  the  solution  to  be 

tested.  Add  about  5  cc.  of  methyl  alcohol  and  2  cc.  of  concen- 
trated sulfuric  acid  and  stir  the  mixture.  If  the  alcohol  is  now 
ignited,  the  flame  will  be  green-bordered.  The  color  is  due  to  the 
formation  of  methyl  borate  ((CH8)8BOg),  which  is  volatile  and 
imparts  a  green  color  to  the  alcohol  flame.  The  alcohol  may  be 
ignited  in  an  evaporating  dish,  or,  better,  put  into  a  test  tube  fitted 
with  a  one-hole  stopper  carrying  a  glass  tip.  The  mixture  in  the 
test  tube  is  heated  and  the  alcohol  ignited  as  ft  issues  from  the  tip. 

SPECIAL  METHOD  FOR  THE  SEPARATION  AND  IDENTIFICA- 
TION OF  METAL  IONS  OF  THE  IRON-ZINC  GROUP  IN  THE 
PRESENCE  OF  PHOSPHATE  OR  OXALATE  IONS 

As  the  phosphates  of  metals  above  the  alkali  group  are 
insoluble  in  water,  if  the  phosphate  ion  is  present  phosphates 
of  the  metals  of  the  iron-zinc  and  alkaline  earth  groups  will 
be  precipitated  when  the  solution  is  made  alkaline  with  am- 
monium hydroxide  before  the  iron-zinc  group  is  precipitated 
with  hydrogen  sulfide  (see  page  56).  Hence,  if  the  phos- 
phate ion  has  been  found,  it  is  necessary  to  use  a  special 
method  for  the  separation  of  these  two  groups.  Tests  should 
always  be  made  for  the  phosphate  ion  before  the  iron-zinc 
group  is  precipitated.  If  it  is  found,  the  filtrate  from  the 
copper  group  is  boiled  with  about  10  cc.  of  bromine  water  to 
oxidize  the  iron  and  a  small  portion  of  this  solution  tested  for 
iron  by  the  usual  reagents  (see  page  56).  Neutralize  the 
remainder  of  the  solution  with  ammonium  hydroxide  and 
acidify  strongly  with  acetic  acid.  Add  5  cc.  of  a  50  per  cent 
ammonium-acetate  solution  and,  if  the  solution  is  not  reddish- 
brown  in  color,  add  ferric  chloride  solution,  drop  by  drop, 
until  this  color  is  produced.  Boil  the  solution,  diluting  with 
water  during  boiling  if  necessary.  As  ferric  phosphate  is  the 
least  soluble  of  any  of  the  phosphates  of  this  group,  this 
procedure  precipitates  all  of  the  phosphate  ion  as  ferric 


PROCEDURE  WITH  AN  UNKNOWN  SUBSTANCE    73 

phosphate,  with  possibly  some  phosphates  of  aluminium  and 
chromium,  while  the  other  metals  remain  in  solution.  Filter, 
and  test  the  filtrate  by  addition  of  more  ammonium  acetate  and 
boil.  Continue  this  treatment  as  long  as  any  ferric  phosphate  is 
precipitated.  When  all  the  phosphate  has  been  removed,  the 
solution  is  treated  by  the  usual  method  for  the  separation  and 
identification  of  the  iron-zinc,  alkaline  earth,  and  alkali  groups. 
If  desired  the  precipitate  may  be  tested  for  aluminium  and 
chromium  by  the  usual  treatment  with  sodium  peroxide. 

METHOD  or  PROCEDURE  WITH  AN  UNKNOWN  SUBSTANCE 

The  substance  will  be  a  liquid  or  a  solid  (gases  will  not 
be  considered  in  this  outline),  and  may  be  either  a  single 
substance  or  a  mixture. 

It  should  first  be  examined  for  the  usual  physical  proper- 
ties,—  particularly  color,  taste  (care!),  odor,  and,  if  it  is  a 
solid,  crystalline  form,  solubility,  etc.  — ,  as  these  may  give 
much  information  in  regard  to  its  nature.  Heating  a  little  of  the 
solid  on  platinum  foil  or,  if  a  liquid,  evaporating  and  heating 
the  residue  will  show  whether  or  not  this  material  is  organic 
and  give  information  as  to  the  melting  and  volatilizing  points. 
In  the  case  of  solids  other  easily  performed  tests,  as  heating 
in  a  closed  tube  and  heating  with  sodium  carbonate  on  char- 
coal, may  frequently  be  used  to  advantage. 

If  the  substance  is  a  solution  it  should  be  tested  with 
litmus,  and  if  neutral  or  acid  the  analysis  begun  without 
further  treatment.  If  alkaline  it  must  be  neutralized  and 
any  precipitate  which  forms  dissolved  before  the  analysis  is 
begun.  When  solids  are  to  be  dissolved,  take  a  small  repre- 
sentative portion  of  the  material  and  try  dissolving  it  in 
distilled  water,  remembering  that  many  substances  are  only 
moderately  soluble  in  water.  If  it  does  not  dissolve,  heat  the 
water.  If  there  is  no  result,  add  a  few  drops  of  dilute  nitric 


74  NOTES  ON  QUALITATIVE  ANALYSIS 

acid,  first  in  the  cold  and  then  with  heat,  gradually  increasing 
the  amount  of  acid  necessary.  If  still  unaffected,  repeat  on 
another  portion,  using  hydrochloric  acid  instead  of  nitric  acid. 
If  both  nitric  and  hydrochloric  acids  fail,  aqua  regia  may  be 
tried,  but  the  instructor  should  first  be  consulted.  When 
large  amounts  of  strong  acids  are  added,  the  excess  of  acid 
should  be  removed  by  evaporating  before  proceeding  with 
the  analysis.  If  the  substance  is  only  partly  soluble  it  is 
sometimes  convenient  to  analyze  the  solution  and  residue 
separately.  Most  residues  insoluble  in  acids  can  be  taken 
into  solution  by  fusing  with  sodium  or  potassium  carbonates 
and  by  subsequent  treatment  with  acids.  In  extreme  cases 
fusion  with  sodium  hydroxide  or  sodium  peroxide  may  be 
necessary.  In  the  case  of  compounds  of  antimony  and  tin 
the  addition  of  sulfur  to  the  fusion  mixture  may  assist. 

PROCEDURE  FOR  ALLOYS 

Take  a  small  portion  of  the  alloy  and  treat  with  dilute 
nitric  acid  with  the  aid  of  heat. 

Most  alloys  will  dissolve  completely  by  this  treatment, 
except  those  containing  antimony  and  tin,  in  which  case  a 
whitish  or  gray  residue  will  remain.  If  gold  or  platinum 
are  in  the  alloy  a  metallic-looking  residue  will  be  left,  while 
carbon,  if  present,  will  appear  as  a  black  residue.  If  the 
residue  is  carbon  it  is  filtered  off  and  discarded,  while  plati- 
num and  gold  are  dissolved  in  aqua. regia. 

The  residue  left  from  alloys  containing  tin  and  antimony 
consists  of  the  oxides  of  these  metals  (SnO2,  Sb0O6).  Treat 
with  hot  concentrated  tartaric  acid  solutions.  Filter  and  test 
the  filtrate  for  antimony  by  acidifying  with  hydrochloric 
acid  and  precipitating  with  hydrogen  sulfide.  The  residue 
may  be  boiled  with  concentrated  hydrochloric  acid  and  this 
solution  tested  for  tin. 


BAKING  POWDEKS  75 

Another  method  consists  in  fusing  the  residue,  of  tin  and 
antimony  oxides,  with  sodium  hydroxide  (or  carbonate)  and 
sulfur,  which  converts  the  oxides  into  the  thio  salts.  The 
fused  mass  is  dissolved  in  water  and  acidified  with  hydrochloric 
cid,  which  precipitates  SnS2  and  Sb2S5.  The  identification  of 
ohese  sulfides  is  then  made  in  the  usual  manner  (see  page  63). 
The  nitric  acid  solution  of  the  alloy  is  evaporated  to  small 
volume  to  remove  excess  of  acid,  water  added,  and  the  anal- 
ysis made  according  to  the  usual  outline  for  the  identification 
of  metal  ions.  In  alloys  it  is  unnecessary  to  test  for  acid  ions 
except  in  rare  cases,  and  tests  for  the  metals  of  the  alkaline 
earth  and  alkali  groups,  with  the  exception  of  magnesium, 
may  be  omitted. 

PROCEDURE  FOR  BAKING  POWDERS 

The  metal  ions  found  in  baking  powders  will  be  found  in 
the  iron-zinc,  alkaline  earth,  and  alkali  groups.  All  salts 
should  be  soluble  in  water.  The  analysis  is  somewhat  inter- 
fered with  by  the  presence  of  the  starch  which  is  in  the 
baking  powder.  In  order  to  get  rid  of  the  starch,  about  5 
grains  of  the  sample  is  stirred  with  50  cc.  of  cold  water  for 
fifteen  to  twenty  minutes.  The  starch  is  allowed  to  settle  and 
the  solution  filtered.  The  filtrate  is  taken  for  analysis,  and 
the  metal  ions  and  acid  ions  are  identified  in  the  usual  way. 


QUESTIONS 


1.  Name  some  physical  properties  frequently  made  use  of  in  iden- 
tifying substances. 

2.  Define  and  illustrate  combination,  decomposition,  double  decom- 
position, reversible  reaction. 

3.  Distinguish  between  mixtures,  compounds,  and  solutions. 

4.  Define  and   illustrate :    solute,  solvent,   dilute   solution,  concen- 
trated solution,  saturated  solution. 

6.  Define    and   illustrate :    gram-molecule,   molar   solution,   normal 
solution,  tenth  normal  solution,  strength  of  solution. 

6.  (1)  Calculate  the  molar  concentration  and  normality  of  a  solu- 
tion   containing   in    100  cc.    36    grams   of    NH4C1;    11.42   grams   of 
(NH4)2CO8-  H2O;  27.02  grams  of  FeCl8  •  6  H2O ;  2  grams  of  CaCl2  •  6  H2O ; 
0.0003  grains  of  BaCrO4 ;    5  grams  of  Ba(OII)2  •  8  II2O ;  4  grams  of 
(NH4)2C2O4  •  H2O  ;    3.58   grams   of   Na2HPO4  •  12  H2O ;    5    grams   of 
K2Cr2O7;     2   grams    of    BaCl2  •  2  H2O.      (2)    What    volume    of    the 
(NH4)2C2O4  solution  would  be  required  to  precipitate  all  the  Ca+  +  ion 
in  5cc.  of  the  CaCl2  solution?   (3)  What  weight  of  BaCrO4  would  In- 
produced  by  precipitating  all  the  Ba+  +  ion  in  5  cc.  of  the  BaCl2  solu- 
tion?  (4)  What  weight  of  MgNH4PO4  could  be  precipitated  by  using 
5  cc.  of  the  Na2HPO4  solution? 

7.  (1)  Using  the  following  data,  plot  the  curves  of  solubility  for 
NaCl,  KNO3,  CaO,  and  MnSO4.    The  solubility  of  each  substance  is 
expressed  as  grams  in  100  grams  of  water. 


T°C 

NaCl 

KN08 

CaO 

MnSO4 

T°C 

NaCl 

KN08 

CaO 

MnSO4 

0 

35.7 

13.3 

0.1382 

55.4 

50 

37.0 

85.9 

74.8 

6 

35.7 

17.2 

58.2 

60 

37.3 

110.9 

0.0869 

65.9 

10 

.35.8 

21.1 

0.1343 

61.1 

65 

37.5 

124.9 

61.5 

15 

35.9 

26.1 

0.1300 

63.8 

80 

38.4 

172.0 

61.5 

20 

36.0 

31.2 

66.3 

90 

39.1 

206.0 

60.3 

30 

36.3 

44.3 

0.1163 

70.4 

100 

39.8 

247.0 

0.0577 

52.9 

40 

36.6 

64.0 

0.1006 

73.1 

76 


QUESTIONS  77 

(2)  At  what  temperature  would  sodium  chloride  and  potassium  nitrate 
have  the  same  solubility?  (3)  In  general,  what  is  the  effect  of  increas- 
ing the  temperature  on  the  solubility  of  substances?  (4)  Discuss  the 
solubility  curves  by  applying  the  law  of  Le  Chatelier. 

8.  The  density  of  a  gas  at  740mm.  pressure  and  20°  is  1.7426. 
What'  is  the  molecular  weight  ?     Analysis  of  the  gas  gave  :   carbon, 
27.27  per  cent ;  oxygen,  72.73  per  cent.    Find  the  molecular  formula. 

9.  How  does   the  strength  of  a   solution  affect  the  boiling   and 
freezing  points  of  the  solution  ?    How  does  it  affect  the  vapor  pressure 
and  osmotic  pressure? 

10.  What  is  the  relation  between  the  molar  concentration  and  the 
factors  mentioned  in  question  9  ? 

11.  State  Avogadro's  hypothesis,  and  explain  or  illustrate  it  so  as  to 
show  that  you  understand  its  meaning. 

12.  If  21.875  grams  of  a  substance  dissolved  in  1000  grams  of  water 
caused  a  lowering  of  the  freezing  point  0.2225°,  what  is  the  molecular 
weight?    If  the  substance  were  composed  of  carbon  39.54  per  cent, 
hydrogen  7.68  per  cent,  and  oxygen  52.78  per  cent,  what  would  be  its 
molecular  formula? 

13.  When  18  grams  of  a  substance  were  dissolved  in  200  grams  of 
water,  the  solution  boiled  at  100.26°.   What  is  the  molecular  weight  ? 

The  percentage   composition  was   carbon,  40  per  cent ;  hydrogen, 
6.7  per  cent;  oxygen,  53.3  per  cent.   Find  the  molecular  formula. 

14.  If  0.00017  grams  of  BaSO4  dissolve  in  100  cc.  of  water,  how  many 
liters  would  be  required  to  dissolve  one  gram?   Such  a  solution  would 
be  saturated  ;  would  it  be  concentrated  ?    If  the   barium   sulf  ate  in 
this  solution  were  100  per  cent  dissociated,  what  would  be  the  molar 
concentration  of  the   Ba  +  +  ion   and  the   solubility-product  constant 
for  BaSO4?     , 

15.  A  mixture  of  NH4C1,  KNO3,  and  Na2SO4  is  dissolved  in  water. 
Make  a  list  of  the  possible  constituents  in  the  solution. 

16.  What  classes  of  substances  when  in  solution  show  abnormal 
changes  in  boiling    and   freezing   point?     How   are   these   abnormal 
changes  explained  ?    What  regularity  in  double  decomposition  reac- 
tions is  shown  by  these  substances  in  solution  which  is  not  shown 
when  they  react  dry? 

17.  Make  a  clear,  brief  statement  of  the  ionic  theory. 

18.  State  several  reasons  for  believing  in  the  ionic  theory. 

19.  What  substances  capable  of  reacting  are  in  each  of  the  following 
solutions :  NaCl,  NaOH,  H2SO4,  CuSO4,  Br2,  Na2HPO4. 


T8  NOTES  ON  QUALITATIVE  ANALYSIS 

20.  How  does  dissociation  change  with  dilution  ?    Does  it  change  in 
the  same  degree  for  every  substance?    Define  strong  acids  and  weak 
acids ;  strong  bases  and  weak  bases.    Name  some  in  each  class. 

21.  Explain  what  we  mean  when  we  say  that  a  solution  in  a  closed 
vessel  does  not  evaporate.    Why  does  a  crystal  of  sodium  chloride  put 
into  a  saturated  solution  of  salt  appear  not  to  dissolve  ? 

22.  Define  equilibrium  as  applied  to  chemical  reactions.    Define  a 
reversible  reaction. 

23.  What  would  be  the  result  of  adding  hydrogen  chloride  to  a 
saturated  solution  of  sodium  chloride?    Explain. 

24.  State  the  Mass  Law  and  illustrate  by  some  examples. 

25.  Using  the  data  given  on  page  5,  calculate  the  ionization  con- 
stants for  HC1,  HC2H3O2,  and  NaCl.    HC1  gas  is  liberated  from  a  con- 
centrated hydrochloric  acid  solution  by  adding  concentrated  sulfuric 
acid.    Explain. 

26.  Using  the  data  given  on  page  10,  calculate  the  molar  concen- 
tration of  OH~  ion  in  0.1  normal  KOH ;  Cl~  ion  in  0.1  normal  HC1 ; 
SO4~  ion  in  0.1  normal  H2SO4 ;  CO8~  ion  in  0.1  normal  H2CO8 ; 
S —  ion  in  0.1  normal  H2S. 

27.  Solid  sodium  acetate  is  added  to  0.01  normal  acetic  acid  until 
the  solution  is  0.1  normal  with  respect  to  the  sodium  acetate.    If  the 
original  solution  were  4.2  per  cent  ionized,  and  the  sodium  acetate 
solution  80  per  cent,  calculate  the  change  in  molar  concentration  of 
the  hydrogen  ion. 

28.  Explain  by  means  of  the  Mass  Law  the  effect  of  adding  NII4C1 
to  a  solution  of  NH4OH.    Why  would  the  effect  not  be  so  noticeable 
if  NaCl  was  added  to  NaOH? 

29.  What  effect  does  the  addition  of  sodium  acetate  have  upon  the 
strength  of  a  solution  of  acetic  acid?    Explain.   Would  the  result  be 
the  same  if  sodium  acetate  was  added  to  a  solution  of  hydrochloric 
acid  ?   Explain. 

30.  (1)  Solid  NH4C1  is  added  to  0.1  molar  NH4OH  solution  until  the 
solution  is  0.1  molar  with  respect  to  the  NH4C1.    If  the  NII4C1  is 
80  per  cent  dissociated,  and  the  original  NH4OH  is  1.3  dissociated,  what 
is  the  change  in  the  molar  concentration  of  the  OH~  ion?    (2)  The 
solubility-product  constant  for  Mg(OH)2  is  3.4  x  10~n.   If  the  above 
NH4OH-NH4C1  solution  were  also  0.1  molar  with  respect  to  MgSO4, 
70  per  cent  dissociated,  would  Mg  (OH)2  be  precipitated  ? 

31.  What  are  the  three  conditions  which  tend  to  bring  reversible 
reactions  to  completion?   Illustrate  each  by  an  equation. 


QUESTIONS  79 

32.  Illustrate  hydrolysis  by  an  equation.  Explain  why  sodium  car- 
bonate reacts  alkaline  in  solution  and  copper  sulfate  reacts  acid  under 
the  same  circumstances. 

•  33.  Is  the  addition  of  water  (or  hydrogen  and  oxygen  in  the  propor- 
tion to  form  water)  oxidation  or  reduction  or  neither  ?    Why  ? 

34.  Name  some  of  the  more  common  reducing  agents  and  tell  why 
each  acts  as  a  reducing  agent. 

35.  Name  some  of  the  more  common  oxidizing  agents  and  tell  why 
each  acts  as  an  oxidizing  agent. 

36.  Define  oxidation  and  reduction  in  terms  of  gain  or  loss  of  hydro- 
gen and  oxygen ;  in  terms  of  gain  or  loss  of  charges. 

37.  Would  the  following  mixtures  of  compounds  be  completely  solu- 
ble in  water,  partly  soluble  in  water  but  completely  soluble  in  acids,  or 
only  partly  soluble  in  both  water  and  acids:  KNO3,  ZnSO4,  A1C13; 
MnCO3,  Zn(NO8)2,  CuSO4;  NaNO3,  K2CO3,  (NH4)2SO4,  Na2HPO4> 
BaSO4,    NaCl;    CaCO3,    FeCl3,    CdCl2,    Hg  (NO3)2 ;    HgCl2,    AsCl3' 
Pb(N03)2,     Co(NO,)g;     Ba(N03)2,     AgNO3,     Pb(NO3)2,     KNO3; 
Ca3(PO4)2,  MnCl2,  NaNO8?, 

38.  Write  a  list  of  the  elements  to  be  studied  in  each  group,  and 
give  group  precipitant  for  each.     Why  is  there  no  group  precipitant 
for  the  alkali  group? 

39.  Why  must  sodium  hydroxide  be  added  before  heating  in  testing 
for  ammonium  ?    Why  cannot  the  litmus  paper  be  put  directly  into  the 
solution  in  making  the  test  ? 

40.  Why  must  the  solution  be  evaporated  to  complete  dryness  and 
heated  strongly  before  testing  for  potassium  with  sodium  cobaltinitrite  ? 
Is  this  procedure  necessary  when  testing  for  potassium  by  the  flame 
test?   Why? 

41.  Why  are  ammonium  salts  so  generally  used  as  reagents  when 
sodium  and  potassium  salts  are  equally  soluble? 

42.  Could  ammonium  be  tested  for  in  the  original  solution  con- 
taining all  the  other  metals  ?    Why  would  it  be  advisable  to  test  for 
it  here? 

43.  Why  do  we  usually  use.  a  hydrochloric  acid  solution  or  moisten 
the  residue  with  hydrochloric  acid  before  testing  in  the  flame  ? 

44.  An  unknown  of  the  alkali  group  is  known  to  contain  only  one 
metal.   What  is  the  first  test  you  would  apply  ?   Would  this  test  identify 
your  metal? 

45.  In  an  alkali  group  unknown,  if  you  had  proved  the  presence  of  am- 
monium and  had  forgotten  to  evaporate  and  ignite  before  precipitating 


80  NOTES  ON  QUALITATIVE  ANALYSIS 

with  sodium  cobaltinitrite,  how  could  you  test  the  precipitate  to  prove 
whether  potassium  is  present  or  not? 

46.  Why   is  ammonium    chloride    added    before    precipitating   the 
alkaline  earth  group  ?    What  element  would  be  difficult  to  identify  if 
the  ammonium  chloride  had  not  been  added  ? 

47.  Could  you  identify  magnesium  in  the  nitrate  from  the  alkaline 
earth  group  by  precipitating  it  with  ammonium  hydroxide  ?   Why  ? 

48.  What  is  the  group  reagent  for  the  alkaline  earth  group?   Why 
do  we  not  use  sodium  carbonate  ? 

49.  Why  do  we  use  acetic  acid  to  dissolve  the  precipitated  carbonates 
of  the  alkaline   earth  group  instead  of  hydrochloric   or  some  other 
strong  acid? 

50.  The  table  on  page  32  shows  that  SrCrO4  is  quite  insoluble.   Why 
is  it  that  barium  can  be  separated  from  strontium  by  means  of  the 
chromate  ion  ?    If   the   chromate  ion  had  been  added  in  a  neutral 
solution  what  would  have  been  the  result? 

51.  Apply  the  Mass  Law  to  the  questions  asked  in  50. 

52.  (1)  Use  the  molar  solubilities  given  on  page  32.    A  saturated 
solution  of  BaC2O4  is  about  86  per  cent  ionized ;  SrC2O4,  89  per  cent ; 
CaC2O4,  96  per  cent;  MgC2O4,  33  per  cent.    Calculate  the  solubility- 
product  constant  for  each  of  these  substances.   (2)  If  0.1  molar  sodium 
oxalate  were  70  per  cent  ionized,  what  would  be  the  least  concentration 
of  calcium  ion  which  could  be  detected  in  such  a  solution  ?   (3)  Would 
the  concentration  of  calcium  ion  in  a  saturated  solution  of  CaSO4,  53 
per  cent  ionized,  be  sufficient  to  give  a  precipitate  of  CaC2O4  ?  (4)  Show 
from  the  data  above  whether  the  oxalate  test  for  calcium  could  be 
used  if  barium,  strontium,  or  magnesium  ions  were  present. 

53.  Why  is  strontium  tested  for  by  calcium  sulfate  and  not  by  sul- 
furic  acid?    Could  potassium  sulfate  have  been  used  under  any  cir- 
cumstances?   Explain.    Apply  the  Mass  Law  to  your  answers. 

54.  Why  should  ammonium  carbonate  and  ammonium  oxalate  be 
added  to  the  nitrate  from  the  alkaline  earth  group  and  the  solution 
filtered  before  testing  for  magnesium  ? 

55.  If  strontium  is  not  completely  removed,  how  and  why  may  it 
interfere  with  the  test  for  calcium  ? 

56.  If  strontium  is  known  to  be  absent,  how  may  the  separation  of 
the  alkaline  earth  elements  be  shortened  ?    How  may  it  be  shortened  if 
barium  is  known  to  be  absent? 

57.  Why  could  we  not  separate  the  alkaline  earth  metals  from  each 
other  as  carbonates  ? 


QUESTIONS  81 

58.  If  a  solution  which  might  contain  alkaline  earth  and  alkali 
metals  gives  no  precipitate  for  an  alkaline  earth  metal  with  (NH4)2CO3, 
would  you  evaporate  to  dryness  and  proceed  at  once  to  test  for  the 
alkali  metal?    Why? 

59.  If  barium  and  strontium  were  both  known  to  be  absent,  could 
you  test  the  solution  containing  alkaline  earth  and  alkali  metals  directly 
with  (NH4)2C2O4  for  calcium  ?    Why  ? 

60.  What  metals  of  the  alkaline  earth  group  can  be  tested  for  by 
the  flame  test  ?  If  a  solution  contained  only  alkaline  earth  and  alkali 
metals,  would  the  flame  test  alone  be  sufficient?    What  metals  might 
still  be  in  doubt?    Could  you  certainly  identify  calcium  or  strontium 
alone  or  in  the  presence  of  each  other  by  the  flame  test  ? 

61.  Would  a  green  flame   always   be  positive  evidence  of  barium? 
What  other  substances  might  cause  a  green  flame? 

62.  List  the  elements  which  are  precipitated  by  hydrogen  sulfide, 
giving  the  color  of  each  precipitate.   Which  are  soluble  in  dilute  acids? 
Which  are  soluble  in  ammonium  sulfide?    Which  are  hydrolyzed  and 
precipitated  as  hydroxides  ? 

63.  In  precipitating  with  hydrogen  sulfide  how  may  we  be  sure  that 
the  precipitating  agent  always    has    the  same  concentration?    What 
is  this  concentration?    Would  you  call  it  a  concentrated  or  a  dilute 
solution? 

64.  Hydrogen  sulfide  is  a  dibasic  acid,  the  principal  ionization  being 
H  +  +  HS~.    Why  do  we  usually  find  normal  sulfides  precipitated  and 
not  acid  sulfides  ?    Why  are  a  few  rnetals  precipitated  as  hydroxides 
and  not  as  sulfides  by  H2S  ? 

65.  If  hydrogen  sulfide  is  passed  into  a  neutral  solution  containing 
chlorides  of  all  the  metals  of  the  copper-arsenic  and  iron-zinc  groups, 
which    sulfides    would    be    completely   precipitated,    which    ones    only 
partially  precipitated,  and  which  ones  would  not  be  precipitated  at  all  ? 
Which  sulfides  would  probably  precipitate  first?    Explain. 

66.  Why  must  the  solution  be  made  distinctly  acid  before  precipi- 
tating the  copper-arsenic  group  with  hydrogen  sulfide  ?     What  would 
be  the  result  if  too  much  acid  is  used  ?   if  too  little  acid  is  used  ? 

67.  Why  can  neither  nitric  acid  nor  sulfuric  acid  be  used  to  acidify 
the  solution  before  precipitating  the  copper-arsenic  group  with  hydrogen 
sulfide  ?   What  acid  is  best  to  use  ? 

68.  Explain  why  arsenic  sulfide  may  be  separated  from  bismuth  sul- 
fide by  means  of  ammonium  sulfide  when  they  both  occur  in  the  same 
group  of  the  periodic  table.    Why  does  stannous  sulfide  require  yellow 


82  NOTES  ON  QUALITATIVE  ANALYSIS 

ammonium  sulfide  to  dissolve  it,  while  stannic  sulfide  is  dissolved  by 
colorless  ammonium  sulfide? 

69.  Why  do  we  obtain  the  sulfides  of  the  metals  when  the  ammo- 
nium sulfide  solution  of  arsenic,  antimony,  and  tin  sulfides  is  treated 
with  a  dilute  acid? 

70.  Write  equations  for  all  the  reactions  included  in  questions  68 
and  69. 

71.  Why  is  it  that  cobalt  and  nickel  sulfides  do  not  precipitate  with 
the  copper-arsenic  group  when  hydrogen  sulfide  is  passed  into  the  acid 
solution,  and  yet  can  be  separated  from  other  members  of  the  iron-zinc 
group  by  treatment  of  the  sulfides   with  hydrochloric  acid? 

72.  Explain  by  the  Mass  Law  why  some  sulfides  are  soluble  in 
hydrochloric  acid  and  others  are  not. 

73.  (1)  Using  the  solubility-product  table  on  pages  42  and  43,  calcu- 
late in  grams  per  liter  the  weight  of  mercuric  ion  (Hg++)  necessary 
to  produce  a  precipitate  of  mercuric  sulfide  when  the  molar  concentra- 
tion of  hydrochloric  acid  present  is  0.2.    (2)  Calculate  in  grams  per 
liter  the  weight  of  lead  ion  (Pb++)  necessary  to  produce  a  precipitate 
of  lead  sulfide  in  the  same  solution.   (3)  In  what  way  would  an  increase 
in  pressure  of  the  hydrogen  sulfide  in  the  precipitation  flask  affect  the 
precipitation  of  metals  as  sulfides  ?   Why  ? 

74.  (1)   WThat   volume   of   hydrogen   sulfide   at   20°  and  740  mm. 
pressure  would  be  required  to  precipitate  completely  the  cadmium  in 
25  cc.  of  a  0.2  molar  solution  of  cadmium  sulfate  ?    (2)  What  weight 
of  iron  sulfide  and  what  volume  of  hydrochloric  acid,  7.29  per  cent  IIC1, 
specific  gravity  1.036,  would  be  required  to  prepare  this  volume  of 
hydrogen  sulfide  f 

75.  Explain  why  aluminium  hydroxide  is  precipitated  from  a  solution 
of  an  aluminium  salt  by  ammonium  hydroxide,  but  when  sodium  hydrox- 
ide is  added  under  similar  conditions  the  precipitate  first  formed  dissolves. 

76.  Explain  why  zinc  hydroxide  is  easily  soluble  in  either  dilute 
hydrochloric  acid  or  strong  sodium  hydroxide. 

77.  Explain  why  zinc  hydroxide  is  completely  soluble  in  an  excess 
of  ammonium   hydroxide,  while  aluminium  hydroxide  is  not  soluble 
«nder  the  same  conditions. 

78.  Supposing  an  excess  of  sodium  hydroxide  were  added  to  a  solu- 
tion containing  all  the  metals  of  the  iron-zinc  group ;  of  what  would 
the  precipitate  consist?    Which  metals  would  remain  in  solution? 

79.  Suppose  in  question  78  ammonium  hydroxide  had  been  added  in- 
stead of  sodium  hydroxide.   Which  metals  would  have  been  precipitated 


QUESTIONS  83 

and  which  would  have  remained  in  solution?    Could  this  be  used  as  a 
means  of  separation? 

80.  What  metals  are  affected,  and  in  what  manner,  when  excess  of 
sodium  peroxide  is  added  to  a  solution  containing  metals  of  the  iron- 
zinc  group  ? 

81.  If  ammonium  sulfide  produces  a  black  precipitate  when  added  to 
a  solution  containing  the  metals  of  the  iron-zinc  group,  what  metals 
might  be  present  ?   If  the  precipitate  is  light-colored  what  metals  might 
be  present  ? 

82.  If  the  ammonium  sulfide  precipitate  of  the  iron-zinc  group  is 
light-colored,  how  may  the  separation  be  shortened?  Outline  the  method. 

83.  If  the  ammonium  sulfide  precipitate  is  pure  white,  how  would 
you  proceed  with  the  analysis  ? 

84.  Given  a  solution  known  to  contain  iron  and  zinc  only,  show  how 
you  would  separate  and  identify  these  two  elements. 

85.  Given  a  solution  known  to  contain  cobalt,  chromium,  and  alumin- 
ium, show  how  you  would  separate  and  identify  each  of  these  elements. 

86.  On  treating  the  dissolved  sulfides  of  the  iron-zinc  group  with 
sodium  peroxide  a  dark-brown  precipitate  and  a  yellow  solution  are 
obtained.    What  two  metals  are  probably  present  ? 

87.  Why  is  it  necessary  to  be  careful  to  decompose  all  the  hydrogen 
peroxide  before  testing  for  zinc  ?    If  in  the  final  test  for  zinc  a  white 
precipitate  insoluble  in  hydrochloric  acid  is  obtained,  would  you  report 
zinc  ?   The  precipitate  probably  is  what  ? 

88.  If  a  phosphate  is  known  to  be  present  in  a  solution,  how  would 
you  modify  the  analysis  for  the  iron-zinc  group  ? 

89.  What  metals  of  the  iron-zinc  group  can  be  identified  by  the 
blowpipe  or  borax-bead  tests  ? 

90.  If  a  precipitate  of  iron,  aluminium,  and  chromium  is  fused  with 
sodium  carbonate  and  potassium  nitrate,  how  would  you  complete  the 
analysis  ? 

91.  If  the  filtrate  from  your  hydrogen  sulfide  precipitate  of  the 
iron-zinc  group  is  colored  pink,  what  is  probably  the  trouble?    How 
would  you  proceed  with  the  analysis  ? 

92.  If  the  solution  formed  by  treating  the  precipitated  sulfides  of 
the  iron-zinc  group  with  dilute  hydrochloric  acid  is  colored  pink,  to  what 
is  the  color  probably  due,  and  what  caused  this  condition?    What 
difficulties  might  it  cause  in  your  analysis? 

93.  If  when  testing  fer  chromium  with  lead  acetate  a  white  pre- 
cipitate is  obtained,  does  it  show  the  presence  of  chromium?    What  is 


84  NOTES  ON  QUALITATIVE  ANALYSIS 

the  precipitate  probably?  How  would  you  prove  the  presence  or  absence 
of  chromium? 

94.  Why  is  the  filtrate  from  the  iron-zinc  group  sometimes  colored 
brown  ?   How  may  this  filtrate  be  cleared  ? 

95.  If  on  making  the  filtrate  from  the  copper-arsenic  group  alkaline 
with  ammenium  hydroxide  a  black  precipitate  is  obtained,  what  does 
it  indicate  ?    What  would  you  do  ? 

96.  If  the  copper-arsenic  group  were  not  completely  removed  by 
hydrogen  sulfide,  what  difficulties  would  it  cause  in  the  analysis,  and 
where  ? 

97.  What  reaction  takes  place  when  borax  is  heated  in  a  flame? 
What  reaction  when  a  metallic  oxide  is  added  to  this  product  ? 

98.  What  is  the  first  thing  you  would  do  if  given  an  unknown 
solution  for  the  copper-arsenic  group?   Why? 

99.  Suppose  on  adding  hydrochloric  acid  to  an  unknown  solution 
for  the  copper-arsenic  group  a  white  precipitate  formed,  what  conclu- 
sions would  you  draw  ?    How  would  you  proceed  with  the  analysis  ? 

100.  If  on  passing  hydrogen  sulfide  into  a  solution  of  the  copper- 
arsenic  group  brown  vapors  are  observed,  and  the  precipitate  formed 
disappears,  or  does  not  increase,  what  conclusions  do  you  draw  ?    How 
would  you  proceed  with  the  analysis  ? 

101.  If  free  nitric  acid  or  much  nitrates  are  present  in  a  copper- 
arsenic  unknown,  what  is  the  result  when  hydrogen  sulfide  is  passed 
into  the  solution?    What  difficulties  will  this  cause  in  the  analysis? 

102.  What  would  be  the  result  in  precipitating  the  copper-arsenic 
group  if  too  little  hydrochloric  acid  is  present?   if  too  much  hydro- 
chloric acid  is  present  ?   What  would  you  do  in  each  case  ? 

103.  If  lead  has  been  identified  in  the  silver  group,  why  should  it  be 
tested  for  and  separated  in  the  copper-arsenic  group  ? 

104.  If  a  copper-arsenic-group  unknown  solution  should  be  treated 
directly  with  nitric  acid  instead  of  first  precipitating  it  with  hydrogen 
sulfide,  and  the  solution  then  treated  with  sulfuric  acid  and  ammonium 
hydroxide,  what  element  would  you  not  have  separated  and  how  would 
it  interfere  with  the  analysis  ? 

105.  If  mercury  has  been  identified  in  the  silver  group,  why  should 
it  be  tested  for  and  separated  in  the  copper-arsenic  group? 

106.  Suppose  colorless  ammonium  sulfide  had  been  used  to  separate 
the  arsenic  group  from  the  copper  group,  what  element  might  there  be 
trouble  in  identifying?    Explain. 

107.  If  a  black  precipitate  is  obtained  by  passing  hydrogen  sulfide 


QUESTIONS  85 

into  a  copper-arsenic  unknown  solution,  what  metals  might  be  present?  If 
the  precipitate  is  yellow  or  orange  yellow,  what  metals  might  be  present  ? 

108.  Suppose  hydrogen  sulfide  produces  a  yellow  precipitate  in  a 
copper-arsenic  unknown,  how  may  the  usual  analysis  be  shortened? 

109.  If  the  precipitate  obtained  in  question  108  is  completely  soluble 
in  yellow  ammonium  sulfide,  how  may  the  analysis  be  shortened?    If 
entirely  insoluble  in  yellow  ammonium  sulfide  how  may  the  analysis 
be  shortened? 

110.  If  the   filtrate  from   the   hydrogen-sulfide   precipitate   of  the 
copper-arsenic    group   is   blue,   what   do   you   suspect?    What  would 
you    do  ? 

111.  If   a  bright-yellow  precipitate   should  be   obtained  when  the 
filtrate  from  the  copper-arsenic  group  is  made  alkaline  with  ammonium 
hydroxide  and  hydrogen  sulfide  added,  what  would  it  signify?    What 
would  you  do  ? 

112.  If  sodium  hydroxide  is  added  instead  of  ammonium  hydroxide 
in  separating  bismuth,  what  other  elements  might  be  precipitated? 
How  would  this  affect  the  rest  of  the  analysis  ? 

113.  Give  two  ways  to  distinguish  between  ferric  and  ferrous  salts ; 
to  distinguish  between  stannic  and  stannous  salts. 

114.  A  solution  containing  ferric  chloride,  potassium   dichromate, 
and  aluminium  chloride  is  made  acid  with  hydrochloric  acid  and  hydro- 
gen sulfide  passed  into  it.  What  is  formed  ?   Give  reactions  to  show  this. 

115.  Sodium  hydroxide  is  added  in  excess  to  a  solution  containing 
ferric  chloride,  potassium  dichromate,  and  aluminium  chloride.    AVhat 
is  formed  ?   Give  reactions. 

116.  Write  equations  showing  the  oxidation  of   ferrous  sulfate  to 
ferric  sulfate  by  means  of  nitric  acid;   the  reduction  of  ferric  sulfate 
to  ferrous  sulfate  by  hydrogen  sulfide. 

117.  If  you  should  forget  to  treat  the  nitric-acid  solution  of  the  sul- 
fides  of  the  copper-arsenic  group  with  sulfuric  acid,  but  proceed  at  once 
to  the  addition  of  ammonium  hydroxide,  what  effect  would  it  have  on 
the  analysis?    How  would  you  complete  the  analysis? 

118.  Explain  why  copper  and  cadmium  are  not  precipitated  with 
bismuth  by  the  addition  of  ammonium  hydroxide.    Why  is  the  solu- 
tion saturated  with  sodium  chloride  before  copper  is  separated  from 
cadmium  by  hydrogen  sulfide  ?    Why  is  the  cadmium  precipitated  on 
diluting  the  solution  after  filtering  off  the  copper  sulfide  ? 

119.  Explain  why  we  must  use  concentrated  hydrochloric  acid  in 
separating  arsenic  from  antimony  and  tin. 


86  NOTES  ON  QUALITATIVE  ANALYSIS 

120.  Sometimes  the  precipitated  sulfides  from  the  yellow  ammonium 
sulfide  solution  are  dark-colored.    What  is  the  cause  of  this  ?  How  may 
it  be  prevented  ? 

121.  How  may  the  white  precipitates  of  bismuth  oxychloride  and 
antimony  oxychloride  be  distinguished? 

122.  After  dissolving  mercuric  sulfide  in  aqua  regia  why  is  it  neces- 
sary to  evaporate  almost  to  dryness  and  take  up  with  water  before 
testing  for  mercury  with  stannous  chloride? 

123.  Why  is  it  that  mercury  and  lead  are  placed  in  both  the  silver 
and  the  copper-arsenic  groups  while  silver,  which  forms  an  insoluble 
precipitate  with  hydrogen  sulfide,  is  placed  only  in  the  former  group  ? 

124.  In  making  the  final  test  for  bismuth,  why  is  it  necessary  to 
dissolve  the  hydroxide  in  the  least  possible  amount  of  acid  and  pour  it 
into  a  large  volume  of  water  ? 

125.  If  metallic  copper  were  placed  in  a  solution  containing  all  the 
metals  of  the  copper-arsenic  group,  what  would  occur  ?   If  metallic  iron 
had  been  placed  in  a  similar  solution,  what  would  occur?   If  platinum 
were  placed  in  the  solution,  what  would  occur  ? 

126.  One  gram  of  pure  iron  wire  is  dissolved  in  sulfuric  acid  and 
diluted  to  250  cc.  How  many  cubic  centimeters  of  potassium  dichromate 
solution  containing  5  grams  per  liter  would  be  required  to  oxidize  the 
ferrous  sulfate  in  50  cc.  of  the  solution,  as  prepared  above,  to  ferric 
sulf  ate  ? 

127.  0.5  gram  of  arsenic  trisulfide  is  dissolved  in  concentrated  hydro- 
chloric acid,  38  per  cent  HC1,  specific  gravity  1.19,  by  adding  a  few 
crystals  of  potassium  chlorate,  as  described  in  Experiment  28.    What 
volume  of  hydrochloric  acid  would  be  required? 

128.  (1)  The  solubility  of  lead  sulfate  at  18  degrees  is  0.041  grams 
per  liter.     If  the  degree  of  ionization  is  92  per  cent,  calculate  the 
solubility-product  constant.    (2)  If  the  solubility  of  lead  chrornate  is 
0.000026  grams  per  liter  and  the  degree  of  ionization  is  100  per  cent, 
what  is  the  solubility-product  constant?    (3)  Could  lead  ion  (Pb++) 
be   precipitated   as   lead   chromate   in   a   saturated    solution   of    lead 
sulfate  ? 

129.  If  a  copper-arsenic-group  unknown  is  known  to  contain  no 
copper,  how  may  the  analysis  be  shortened  ?    If  it  is  known  to  contain 
only  copper-group  metals,  how  may  it  be  shortened  ?    If  it  is  known  to 
contain  no  lead,  how  may  it  be  shortened  ? 

130.  An  unknown  solution  is  known  to  contain  only  one  member  of 
the  arsenic  group.    Could  the  analysis  be  shortened  ?    How  ? 


QUESTIONS  87 

131.  A  solution  known  to  contain  only  one  element  gives  a  precipi- 
tate with  sodium  hydroxide  which  is  soluble  in  an  excess  of  the  reagent. 
How  would  you  proceed  with  the  analysis? 

132.  If  the  residue  left  when  the  copper-group  sulfides  are  treated 
with  nitric-acid  is  not  soluble  in  aqua  regia,  can  it  be  mercury?    What 
is  it,  probably?    If   the   nitric-acid  residue  is  not  black,  can  it  be 
mercury  ? 

133.  If  an  unknown  solution  is  alkaline,  what  substances  might  be 
precipitated  upon  the  addition  of  hydrochloric  acid  to  precipitate  the 
silver  group?    Why  do  we  first  acidify  with  nitric  acid?    Why  could 
sulfuric  acid  not  be  used? 

134.  An  unknown  solution,  known  to  contain  no  members  of  the 
silver  group,  gives  a  white  precipitate  when  hydrochloric  acid  is  added. 
What  do  you  suspect  the  precipitate  to  be  ?  Would  you  have  to  modify 
your  method  of  analysis  ? 

135.  If  considerable  nitric  acid  is  added  to  a  silver-group  unknown 
solution,  and  then  an  excess  of  hydrochloric  acid,  and  the  solution 
boiled,  what  effect  may  it  have  on  some  of  the  metals  ?    How  would  you 
test  for  the  elements  so  effected? 

136.  Why  is  it  best  to  precipitate  the  silver  group  from  the  cold 
solution  ?    If  you  heated  the  solution,  what  element  would  not  precipi- 
tate ?    Where  would  you  find  it  ? 

137.  If  you  added  ammonium  hydroxide  to  a  silver-group  precipitate 
before  washing  out  the  lead  chloride,  what  would  be  the  result  ?    What 
would  you  do  to  complete  the  analysis  ? 

138.  A  saturated  solution  of  silver  chloride  is  made  0.1  molar  with 
respect  to  sodium  chloride.    Assume  the  sodium  chloride  85  per  cent 
ionized  and  the  silver  chloride  100  per  cent.    If  the  solubility  of  silver 
chloride  is  1.5  x  10  ~4  grams  per  liter,  would  the  silver  chloride  pre- 
cipitate when  the  sodium  chloride  is  added,  and  what  change  in  the 
concentration  of  the  silver  ion  would  occur? 

139.  (1)  The  solubility  of  silver  sulfide  is  2  x  10~5  grams  per  liter. 
Assume  the  silver  sulfide  in  a  saturated  solution  100  per  cent  ionized, 
and  calculate  the  solubility-product  constant.    (2)  By  comparing  with 
the  solubility-product  constant  of  silver  chloride,  determine  what  con- 
centration of  chlorine  ion  (Cl  -)  would  be  required  to  precipitate  silver 
chloride  from  a  saturated  solution  of  silver  sulfide  ? 

140.  Using  the  table  on  page  43,  determine  the  ratio  of  the  con- 
centrations of  the  sulfide  ion  present  in  saturated  solutions  of  copper 
sulfide  and  zinc  sulfide.   Assume  both  sulfides  completely  ionized. 


88  NOTES  ON  QUALITATIVE  ANALYSIS 

141.  During  analysis  one  gram  of  the  sulfides  of  the  copper  group 
was  obtained,  containing  40  per  cent  lead  sulfide,  30  per  cent  copper 
sulfide,  and  30  per  cent  -mercuric  sulfide.    How  many  cubic  centimeters 
of  dilute  nitric  acid,  20  per  cent  HNO3,  specific  gravity  1.1,  will  be  re- 
quired to  dissolve  the  soluble  sulfides  ? 

142.  During  analysis  2  grams  of  precipitate  for  the  iron-zinc  group 
was  obtained,  containing  30  per  cent  chromium  hydroxide,  30  per  cent 
ferrous  sulfide,  30  per  cent  manganese  sulfide,  and  10  per  cent  nickel 
sulfide.   (1)  What  volume  of  dilute  hydrochloric  acid,  7.29  per  cent  HC1, 
specific  gravity  1.036,  would  be  required  to  dissolve  the  soluble  part 
of  the  precipitate?    (2)  What  weight  of  sodium  peroxide  would  be 
required  to  change  the  chromium  chloride  formed  in  (1)  to  sodium 
chromate  ?  (Only  that  part  of  the  sodium  peroxide  actually  taking  part 
in  the  reaction  is  considered.)   (3)  If  an  excess  of  sodium  peroxide  were 
used  in  (2),  what  weight  of  ferric  hydroxide  would  be  formed?    (4)  If 
the  analysis  were  continued  as  directed  in  the  outline  (p.  56  ),  what 
weight  of  permanganic  acid  (HMnO4)  would  be  in  the  solution  which 
forms  the  final  test  for  manganese? 

143.  Given  a  solution  containing  silver,  copper,  and  zinc,  outline 
how  you  separate  and  identify  each  metal. 

144.  Given  a  solution  containing  arsenic,  bismuth,  iron,  and  barium, 
outline  how  you  would  separate  and  identify  each  metal. 

145.  A  known  solution  contains  lead,  chromium,  cobalt,  and  calcium. 
Show  how  you  would  separate  and  identify  each  metal. 

146.  Given  a  solution  containing  cadmium,  zinc,  aluminium,  and 
magnesium,  show  how  you  would  separate  and  identify  each  element. 

147.  Given  a  solution  containing  tin,  mercury  (mercuric),  manga- 
nese, and  ammonium,  outline  how  you  would  separate  and  identify 
each  element. 

148.  Why  is  sodium  carbonate  added  to  a  solution  before  testing 
for  the  acid  ions?   What  tests  might  be  interfered  with  if  this  were 
not  done? 

149.  Why  should  the  filtrate  from  the  sodium-carbonate  precipitate 
be  neutralized  before  testing  for  sulfuric  acid  with  barium  chloride  ? 

150.  If  a  solid  substance  gives  a  gas  when  treated  with  dilute  sul- 
furic acid,  what  acids  might  be  present?    If  the  gas  is  odorless  and 
colorless,  what  acid  might  be  present?    How  would  you  prove  it? 

151.  If  a  solution  gives  no  precipitate  with  magnesium  mixture, 
what  acids  are  absent? 


QUESTIONS  89 

152.  A  water  solution  of  an  unknown  substance  is  found  to  contain 
silver,  copper,  zinc,  and  barium ;  what  acids  cannot  be  present  ?_ 

153.  A  water  solution  of  an  unknown  substance  is  found  to  contain 
the  sulf ate  ion ;  what  metals  cannot  be  present  ? 

154.  An  unknown  solid  is  found  to  contain  the  carbonate  ion.    If  it 
is  completely  soluble  in  water,  what  metals  are  absent? 

155.  A  water  solution  of  an  unknown  substance  contains  the  phos- 
phate ion ;  what  metals  cannot  be  present  ? 

156.  If  analysis  shows  that  an  unknown  substance  which  is  com- 
pletely soluble  in  water  contains  iron,  aluminium,  zinc,  strontium,  and 
potassium,  what  acid  ions  must  be  absent? 

157.  An  unknown  solution  which  contains  a  sulfate  is  insoluble  in 
water,  but  dissolves  in  dilute  hydrochloric  acid ;  what  metals  are  sure 
to  be  absent  ? 

158.  If  you  were  given  a  solid  unknown,  how  would  you  proceed 
with  it  to  get  ready  to  begin  the  analysis  ? 

159.  Given  a  solid  metal  or  alloy,  how  would  you  proceed  with  it  ? 

160.  What  acid  would  you  use  to  dissolve  an  alloy  which  is  given  you 
for  analysis?    If  you  used   hydrochloric  acid  what  difficulties  might 
arise  ? 

161.  An  unknown  which  is  given  you  is  insoluble,  or  only  partially 
soluble,  in  water  ;  how  would  you  proceed  to  get  it  into  solution  ? 

162.  Why  is  the  following  report  on  an  unknown  evidently  wrong? 
White  solid,  partially  soluble  in  water,  completely  soluble  on  the  addi- 
tion of  a  few  drops  of  dilute  hydrochloric  acid.    Found  lead,  bismuth, 
aluminium,  zinc,  barium,  potassium,  chloride,  sulfate,  acetate. 


INDEX 


Acid  ions,  69,  71 

Acidity,  testing  for,  25  ;  methyl  vio- 
let test,  48 

Alkali  group,  30  ;  analysis  for,  38 

Alkaline  earth  group,  31  ;  analysis 
for,  38 

Alkalinity,  testing  for,  25 

Alloys,  procedure  for  analysis,  74 

Aluminium,  29,  40,  49,  55 ;  analysis 
for,  56 

Ammonium,  test  for,  30 ;  analysis 
for,  38 

Ammonium  chloride,  effect  on  solu- 
tion of  magnesium  hydroxide, 
32,  55 

Ammonium  hydroxide,  degree  of 
ionization,  10 ;  ionization  constant, 
13  ;  complex  ions  of,  50 

Amphoteric  hydroxides,  49,  52 

Analysis,  of  the  alkali  and  the  alka- 
line earth  groups,  38  ;  of  the  iron- 
zinc  group,  56  ;  of  the  copper  group, 
61 ;  of  the  arsenic  group,  63 ;  of  the 
silver  group,  68  ;  of  an  unknown 
substance,  73-75;  separation  of  the 
arsenic  and  copper  groups,  65 

Antimony,  29  ;  thio  acids  of,  46,  47  ; 
analysis  for,  63  ;  separation  from 
tin  and  test  for,  64 

Arsenic,  29 ;  thio  acids  of,  46,  47  ; 
analysis  for,  63,  64 

Arsenic  group,  theory  of  separation 
from  copper  group,  45 ;  experi- 
ments, 47,  63,  64  ;  analysis,  63,  65 

Avogadro's  hypothesis,  4 


Baking  powders,  76 

Barium,  30,  31  ;  precipitation  as 
chromate,  34 ;  separation  from 
strontium,  35  ;  analysis  for,  38 

Bismuth,  29,  40,  47,  59  ;  test  for,  60 ; 
analysis  for,  61 

Boiling  point,  elevation  3 

Borax  bead,  54,  58 

Calcium,  30,  31,  35  ;  analysis  for,  38 
Chromates,    action    with    hydrogen 

sulfide,  53;    color  of  ion,  53,   54 
Chromium,  29,   40,  47,   52,   53,   55; 

analysis  for,  56 
Cobalt,  29,  40,  47,  52,  58  ;  analysis 

for,  56 

Complex  ions,  50,  66,  67 
Concentration,  11 
Copper,  29,  40,  47,  59,  60 ;  analysis 

for,  61 
Copper  group,  29,  59 ;  analysis  of,  61 ; 

separation  from  arsenic  group,  65 
Cyanides,  complex,  50 

Decantation,  27 

Dissociation,  5  ;  constant  of,  12 

Electromotive  series,  21 

Electron  theory,  20     * 

Equilibrium,  5-8,  12-16 

Experiments,  alkali  group  (1-4),  30, 
31 ;  alkaline  earth  group  (5-10), 
36,  37 ;  separating  sulfides  into 
groups  (11-13),  47,  48;  iron-zinc 
group  (14-21),  55,  56,  58  ;  copper 


92 


NOTES  ON  QUALITATIVE  ANALYSIS- 


group  (22-26),  60  ;  arsenic  group       Manganese  hydroxide,  52,  53 
(27-29),  64  ;  silver  group  (30-33),       Mass  Law,  9 
67,  68 ;  acid  ions,  71,  72 


Filtration,  26 

Flame  tests,  31,  35,  37 

Freezing  point,  lowering,  3 


Mercury    (ic),    29,    40,    47,    59,   60; 

analysis  for,  61 
Mercury  (ous),  29,  66 
Methyl  violet,  test  for  acidity,  48 

Nickel,  29,  40,  47,  51 ;  analysis  for, 
56 ;  tests  for,  58 


General  instructions,  24-^29 
Groups,  29 

Oxidation,  20,  22,  24,  53,  67 
Hydrogen  sulfide,  use  of,  27 ;  con- 
stants for,  41 ;  table,  42  Potassium,  30  ;  analysis  for,  38 
Hydrolysis,  18,  40  Precipitation,  26 

Reactions,  reversible,  6;  completion 


of,  15;  oxidation  and  reduction, 
20-24 

Reduction,  20-24 ;  of  chromates,  53 ; 
of  antimony  compounds,  62,  64 


lonization,  9 ;  table,  10 
.lonization  constant,  12 ;  for  ammo- 
nia,  13 ;    effect  of  common  ion, 
14,  32,  42 

lonogen,  4 

Ions,  4,  21 

Iron,  29,   40,   47,   50,  53 ;    analysis      Silver,  29,  66,  68 

for,  56  Sodium,  30;  analysis  for,  38 

Iron-zinc  group,  29,  49,  52 ;  oxida-      Solubility,  24,  32 

tion   reactions  for,    53;    analysis      Solubility  product,  17,  32;  table,  43 
for,  56;  special  method  for,  72          Solutions,  2-4,  70 

Strontium,  30,  31,  35,  38 

Lead,  29,  40,  44,  47,  59,  60 ;  analysis      Sulfide  groups,  40-46 
for,  61,  66  ;  in  silver  group,  68  Sulfo  acids  (see  Thio  acids) 

Magnesium,  30,  31 ;  analysis  for,  38  Thio  acids,  46,  75 

Magnesium  carbonate,  solubility  in  Tin,  29,  40,  47  ;  analysis  for,  63 

ammonium  salts,  33 

Manganese,  29,  40,  47,  53  ;  analysis  Zinc,  29,  40,  47,  52  ;   analysis  for, 


for,  66 ;  test  for,  58 


56,  57 


M123271 


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THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
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OVERDUE. 

N     THE    SEVENTH     DAY             
OMIC 

;IGHT 

Alumin 
Antimo 

APR  21  1942  P 

•AY  11  IM7  2ft       6.0 

Argon 
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0.2 

Barium 
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JUN    3    1348 

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Mercury 

Hg  '       200.6 

1 

